A certain drug is used to treat asthma. In a clinical trial of the drug, 27 of 273 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 9% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below. 1-PropZTest prop < 0.09 z-0.513905185 p-0.6963408589 p-0.0989010989 n= 273 a. Is the test two-tailed, left-tailed, or right-tailed? O Two-tailed test Left-tailed test O Right tailed test b. What is the test statistic? z= (Round to two decimal places as needed.) c. What is the P-value? P-value = (Round to four decimal places as needed.)

A certain drug is used to treat asthma. In a clinical trial of the drug, 27 of 273 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 9% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below. 1-PropZTest prop < 0.09 z-0.513905185 p-0.6963408589 p-0.0989010989 n= 273 a. Is the test two-tailed, left-tailed, or right-tailed? O Two-tailed test Left-tailed test O Right tailed test b. What is the test statistic? z= (Round to two decimal places as needed.) c. What is the P-value? P-value = (Round to four decimal places as needed.)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

(5). Calculate SS, variance and standard deviation for the following sample of n=7 scores: 8, 6, 5, 2, 6, 3, 5.
Two accounting professors decided to test if the variance of their grades is different or no. To accomplish this, they each graded the same 10 exams with the following results: Standard Deviation 22.4 Mean Grade Professor 1 79.3 Professor 2 82.1 12.0 F What is the P-value of the test? Select one: a. between 0.01 and 0.025 Ob. between 0.025 and 0.05 C. between 0.05 and 0.10 Od. bigger than 0.10
In a random sample of 5 mobile devices, the mean repair cost was $60.00 and the standard deviation was $12.00. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 95% confidence interval for the population mean. Interpret the results.
A random sample of n =18 data from a normal distrinution with unknown variance = 25.1 and s=3. 24. a. Find a 98% confidence interval for the population mean. b. Interpret the interval.
A 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is less than 0.5. Assume that sample data consists of 78 girls in 169 births, so the sample statistic of 613 results in a z score that is 1 standard deviation below 0. Complete parts (a) through (h) below.Click here to view page 1 of the Normal table. LOADING... Click here to view page 2 of the Normal table. LOADING...Question content area bottomPart 1a. Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below.A.H0: p≠0.5H1: p<0.5B.H0: p=0.5H1: p≠0.5C.H0: p=0.5H1: p<0.5D.H0: p=0.5H1: p>0.5Part 2b. What is the value of α?α=enter your response here (Type an integer or a decimal.)Part 3c. What is the sampling distribution of the sample statistic? Normal distribution Student (t) distribution χ2Part 4d. Is the test two-tailed, left-tailed, or right-tailed? Right-tailed…
A 0.01 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is different from 0.5. Assume that sample data consists of 55 girls in 100 births, so the sample statistic of 1120 results in a z score that is 1 standard deviation above 0. Complete parts (a) through (h) below.Click here to view page 1 of the Normal table. LOADING... Click here to view page 2 of the Normal table. LOADING...Question content area bottomPart 1a. Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below.A.H0: p=0.5H1: p≠0.5B.H0: p=0.5H1: p>0.5C.H0: p=0.5H1: p<0.5D.H0: p≠0.5H1: p=0.5Part 2b. What is the value of α?α=enter your response here (Type an integer or a decimal.)Part 3c. What is the sampling distribution of the sample statistic? χ2 Normal distribution Student (t) distribution
In a random sample of 17 people, the mean commute time to work was 30.2 minutes and the standard deviation was 7.1 minutes. Assume the population is normally distributed and use a t- distribution to construct a 95% confidence interval for the population mean y. What is the margin of error of y? Interpret the results.
Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"
2. A common claim is that garlic lowers cholesterol levels. In a test of the effectiveness of garlic, 49 subjects were treated with doses of raw garlic, and their cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol have a mean of 0.4 and a standard deviation of 21. Use the sample n = 49 to construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. n=49₁ x=0,4₁ 5=21₁ CL = 95%, df=n-1 df=49-1=48
You conduct a between-subjects ANOVA investigating whether different states produce tastier cheeses. You find that eta-squared is .50. What does this tell you? a. The effect size is small. b. 50% of the total variance in tastiness is due to differences between the states. c. The difference between the group means is .50 standard deviations. d. All of the above
A 0.05 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is greater than 0.5. Assume that sample data consists of 78 girls in 144 births, so the sample statistic of 1324 results in a z score that is 1 standard deviation above 0. Complete parts (a) through (h) below.Click here to view page 1 of the Normal table. LOADING... Click here to view page 2 of the Normal table. LOADING...Question content area bottomPart 1a. Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below.Part 4d. Is the test two-tailed, left-tailed, or right-tailed? Left-tailed Right-tailed Two-tailedPart 5e. What is the value of the test statistic?The test statistic is enter your response here. (Type an integer or a decimal.)Part 6f. What is the P-value?The P-value is enter your response here. (Round to four decimal places as needed.)
The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines are randomly assigned to be tested. Both populations have normal distributions. Table 10.4 shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance. Engine Sample Mean Number of RPM Population Standard Deviation 1 1,500 50 2 1,600 60