A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. Thehardness of a sample from each batch of tablets produced is measured to control the compression process. The target value for11.5. The hardness data for a random sample of 20 tablets from one large batch are given.the hardness isН11.62711.37411.38311.477=11.61311.59211.71511.570A hypothesis test ofHo: μ = 11.2Ha: μ # 11.211.49311.45811.48511.62311.60211.55211.50911.47211.36011.46311.42911.531where μ = the true mean hardness of the tablets using a = 0.05 has a P-value of 0.4494. Because the P-value of0.4494 > a = = 0.05, we fail to reject Ho. We do not have convincing evidence that the true mean hardness of these tablets isdifferent from 11.5.A 95% confidence interval for the true mean hardness measurement for this type of pill is (11.472, 11.561).Which is the following statements is not true with regards to the 95% confidence interval and the conclusion from thehypothesis test?The 95% confidence interval provided gives an approximate set of μ's that would not be rejected by a two-sided test atthe α = 0.05 significance level.A two-sided test only allows us to reject (or fail to reject) a hypothesized value for a particular population parameter.Based upon the confidence interval we would not reject μ = 11.5 at the a = 0.05 significance level.The value 11.5 is a plausible value because it falls within the 95% confidence interval.All of the statements are true.

Given, that it fails to reject the null hypothesis, using a p-value approach and the 95% confidence…

A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various illers. hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value f the hardness is μ = 11.5. The hardness data for a random sample of 20 tablets from one large batch are given. 11.627 11.374 11.383 11.477 11.613 11.592 11.715 11.570 A hypothesis test of Ho: μ = 11.2 Ha: 11.2 11.493 11.458 11.485 11.623 11.602 11.552 11.509 11.472 11.360 11.463 11.429 11.531 where the true mean hardness of the tablets using a = 0.05 has a P-value of 0.4494. Because the P-value of 0.4494 > α = 0.05, we fail to reject Ho. We do not have convincing evidence that the true mean hardness of these tablets is

A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various illers. hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value f the hardness is μ = 11.5. The hardness data for a random sample of 20 tablets from one large batch are given. 11.627 11.374 11.383 11.477 11.613 11.592 11.715 11.570 A hypothesis test of Ho: μ = 11.2 Ha: 11.2 11.493 11.458 11.485 11.623 11.602 11.552 11.509 11.472 11.360 11.463 11.429 11.531 where the true mean hardness of the tablets using a = 0.05 has a P-value of 0.4494. Because the P-value of 0.4494 > α = 0.05, we fail to reject Ho. We do not have convincing evidence that the true mean hardness of these tablets is

College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter4: Polynomial And Rational Functions

Section4.6: Rational Functions

Problem 11SC: Find the mean hourly cost when the cell phone described above is used for 240 minutes.

See similar textbooks

Related questions

Q: A short column has a diameter X₁, and is loaded with an axial compressive load X₂. The ultimate…

Q: Samples are collected from the River X and pH values are observed. 9 different pH values are listed…

A: Given data, pH values of river X. Concentrations 4.2, 3.8, 6, 3.5, 8.1, 4.9, 4.1, 3.2, 4. Arranging…

Q: Q3. Milk is filled into 250mL and 1L cartons at a creamery. 10 randomly selected cartons are…

A: The question is about measures.Given :To find :Compare the dispersion using the coefficient of…

Q: Let ?1 denote true average tread life for a premium brand of P205/65R15 radial tire, and let ?2…

A: Answer Given Sample mean [x] =42100 Sample mean [y] =36200 Sample standard deviation [s1] =2400…

Q: Find Upper and lower 95% CI for all Need asap...thanks

A: Introduction - Confidence interval CI=x±tsnwhere , x=mean t=confidence interval value s=standard…

Q: An engineer in charge of a production process that produces a stain for outdoor decks has designed a…

A: Given Information: Population mean (u) = 40 Sample mean (y) = 36.5 Standard Deviation (S) = 13.1…

Q: "Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing.…

A: Obtain the 90% confidence interval for the difference between two means. The 90% confidence interval…

Q: The "spring-like effect" in a golf club could be determined by measuring the coefficient of…

A: From the provided information, Sample size (n1) = 12 and n2 = 12 Level of significance (α) = 0.05

Q: The average cost of food delivered by Talabat is 20 Oman Rials per delivery. The variance of the…

A: Given: Cost of food = X average of X = 20 Oman Rials Variance of X = 16 Oman Rials Standard…

Q: The data below give the concentration of thiol (NM) in the blood lysate for two groups of…

A: Provided sample observations: Normal Rheumatoid 1.85 2.81 1.9 2.85 2.35 4.08 1.45 3.95…

Q: LAB 1: Descriptive Statistics x + cure.com/courses/95110/quizzes/555322?module_item_id%3D3845398 013…

Q: Hydrogen content is conjectured to be an important factor in porosity of aluminum alloy castings. An…

A: a) Consider that ρ is the population correlation coefficient between content (x) and gas porosity…

Q: Total plasma volume is important in determining the required plasma component in blood replacement…

A: (a) Obtain the critical value at the 99% confidence interval/ The critical value is obtained…

Q: Trace metals in drinking water affect the flavor and an unusually high concentration can pose a…

A: Bottom 0.43 0.266 0.567 0.531 0.707 0.716 0.651 0.589 0.469 0.723 Surface 0.415 0.238 0.39 0.41…

Q: (c) Assume th at the resist ance of a cert ain type of resistors is normally distributed with a mean…

Q: "Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing.…

A: Given: Permafresh:, n1 = 7 Hylite: n2 = 4 Formula Used: Confidence Interval = X1 -X2 ± t*×sp×1n1+1n2…

Q: Total plasma volume is important in determining the required plasma component in blood replacement…

A: Hey, since there are multiple subparts posted, we will answer first three question. If you want any…

Question

A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The
hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value for
11.5. The hardness data for a random sample of 20 tablets from one large batch are given.
the hardness is
Н
11.627
11.374
11.383
11.477
=
11.613
11.592
11.715
11.570
A hypothesis test of
Ho: μ = 11.2
Ha: μ # 11.2
11.493
11.458
11.485
11.623
11.602
11.552
11.509
11.472
11.360
11.463
11.429
11.531
where μ = the true mean hardness of the tablets using a = 0.05 has a P-value of 0.4494. Because the P-value of
0.4494 > a = = 0.05, we fail to reject Ho. We do not have convincing evidence that the true mean hardness of these tablets is
different from 11.5.
A 95% confidence interval for the true mean hardness measurement for this type of pill is (11.472, 11.561).
Which is the following statements is not true with regards to the 95% confidence interval and the conclusion from the
hypothesis test?
The 95% confidence interval provided gives an approximate set of μ's that would not be rejected by a two-sided test at
the α = 0.05 significance level.
A two-sided test only allows us to reject (or fail to reject) a hypothesized value for a particular population parameter.
Based upon the confidence interval we would not reject μ = 11.5 at the a = 0.05 significance level.
The value 11.5 is a plausible value because it falls within the 95% confidence interval.
All of the statements are true.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Similar questions

Find the mean hourly cost when the cell phone described above is used for 240 minutes.
A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed of the film, and believes that this can be achieved by reducing the thickness of the film to 20 mils. Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured. For the 25-mil film, the sample data result is = 1.15 and 81 = 0.11, while for the 20-mil film, the data yield 2 = 1.06 and 82 = 0.09. Note that an increase in film speed vould lower the value of the observation in microjoules per square inch. (a) Do the data support the claim that reducing the film thickness increases the mean speed of the film? Use a = 0.10 and assume that the two population variances are equal and the underlying population of film speed is normally distributed. What is the P-value for this test? Round your answer to three decimal places (e.g. 98.765). The data the claim that reducing the film…
Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05 H0: μd=0 H1: μd≠0 T statistic: t = 1.053, P value = 0.327 Degrees of Freedom n-1 = 8-1= 7 critical value 2.3646 If t is less than 2.3646, or greater than 2.3646, reject the null hypothesis Level of significance: α=0.05 question: what a type 1 error and type 2 error would mean. Is it possible that we could have committed a type 2 error in conducting the test
Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05. Q) If a participant has the same corneal thickness in their normal eye as the eye with Glaucoma, what would be the value for difference: measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma.
Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05. Q)Conduct a hypothesis test to determine if there is sufficient evidence to conclude that corneal thickness is different in normal eyes compared to eyes with glaucoma? Write up your results using the 8 steps.
Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05. Hypothesis: H0: μd=0Ha: μd≠0 Using output Test statistics : t=0.134P value=0.897 Degrees of freedom (df): df=7 Level of significance: α=0.05 Decision: P value > 0.05 thus we fails to reject null hypothesis. Question a)Write a report summarizing your findings. When writing the report consider that medical staff estimate that a difference of 4.5 microns or more could impact on their ability to interpret eye pressure correctly. b) Define for the hypothesis stated in part b) what a type 1 error and type 2 error would mean. Is it possible…
The compressive strength, in kilopascals, was measured for concrete blocks from five different batches of concrete, both three and six days after pouring. The data are as follows. Can you conclude that the mean strength after three days is greater than the mean strength after six days? Let μ1 represent the mean strength after three days and μd = μ1 - μ2. Use the a = 0.01 level and the P-value method with the table. Block 1 2 3 4 5 After 3 days 1389 1380 1302 1377 1336 After 6 days 1314 1321 1318 1386 1356 Part (a) State the appropriate null and alternate hypotheses. H0: H1: This is a ___ test. Part (b); _ < P-value <= _ Part (c);There __ enough evidence to conclude that the mean strength after three days differs from the mean strength after six days.(is/is not)
A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period. Single Policyholders Married Policyholders ni = 412 n2 = 778 Number making claims = 75 Number making claims = 111 a. Use a= 0.05. Test to determine whether the claim rates differ between single and married male policyholders. z-value (to 2 decimals) p-value (to 4 decimals) We cannot conclude that there is the difference between claim rates. b. Provide a 95% confidence interval (to 4 decimals) for the difference between the proportions for the two populations. Enter negative answer as negative number.
I uploaded the data for 7.71 but, I need 7.86. For 7.71 my answers were, test stat = 1.6160, D.F= 400, Pvalue=.1081, C.I.=(-.375,3.775) and I concluded that we fail to reject H0
Help 16.
A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed of the film and believes that this can be achieved by reducing the thickness of the film to 20 mils. Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured. For the 25-mil film, the sample data result is.x₁ = 1.15 and S₁ = 0.11, while for the 20-mil film, the data yield 2 = 1.06 and s2 = 0.09. Note that an increase in film speed would lower the value of the observation in microjoules per square inch. Do the data support the claim that reducing the film thickness increases the mean speed of the film? Use a = 0.10 and assume that the two population variances are equal and the underlying population of film speed is normally distributed. The appropriate decision for the test is to reject the null hypothesis True False
A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period. Z-value p-value Single Policyholders n₁ = 440 Number making claims = 88 a. Use a = 0.05. Test to determine whether the claim rates differ between single and married male policyholders. (to 2 decimals) (to 4 decimals) X Married Policyholders n2 = 840 Number making claims 126 We can conclude ✔ that there is the difference between claim rates. b. Provide a 95% confidence interval (to 4 decimals) for the difference between the proportions for the two populations. Enter negative answer as negative number. X * )