DO2WSXandU(Round to four decimal places as needed.)d. What is the null hypothesis, and what do you conclude about it?Identify the null hypothesis.OA. Ho: p=0.11OB. Ho: P# 0.11OC. Ho: p>0.11OD. Ho: p<0.111273Decide whether to reject the null hypothesis. Choose the correct answer below.OA. Reject the null hypothesis because the P-value is greater than the significance level, a.OB. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a.OC. Fail to reject the null hypothesis because the P-value is greater than the significance level, a.#3ED80F3C-$4FARFAPR6aV%5♥FATG6BtvMacBook AirYH&7NUCJ8AMDIIFOI(9K820DF9двO<I)-0LcommandW4F 10PAN--:I;diF11option{[+=?1111 A certain drug is used to treat asthma. In a clinical trial of the drug, 28 of 278 treated subjects experienced headaches (based on data from the manufacturer). Theaccompanying calculator display shows results from a test of the claim that less than 11% of treated subjects experienced headaches. Use the normal distribution as anapproximation to the binomial distribution and assume a 0.01 significance level to complete parts (a) through (e) below.a. Is the test two-tailed, left-tailed, or right-tailed?O Two-tailed testO Right tailed testO Left-tailed testb. What is the test statistic?1273(Round to two decimal places as needed.)c. What is the P-value?P-value=(Round to four decimal places as needed.)d. What is the null hypothesis, and what do you conclude about it?Identify the null hypothesis.#3ETToD80F3с$RFAPR6aV%5TGB6dtv ♫MacBook AirYH&7SZACN8UJ+8MFOIK(9H<I8DDF9OL)OcommandP>F10.I-:;option4.F11{[?1+="1I1-Prop2Testprop<0.11z=-0.494545167p=0.3104606081p=0.100719424545Sawn=2781Question ViewerNextdeletereturnsk

Let x= The number of clinical trial of the drug treated subjects experianced headaches. x=28 n=Total…

Answered: A certain drug is used to treat asthma.…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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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
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…
. 3. A researcher is interested in knowing if preterm infants with late metabolic acidosis and preterm infants without the condition differ with respect to urine levels of a certain chemical. The mean levels, standard deviations, and sample sizes from the two samples are given below. Assuming that the populations have equal variances: a. Fill in the missing pieces of the Stata output. ttesti 38 8.5 3.5 32 6.85 2.6 X y Two-sample t test with A variances combined diff Obs Ho: diff=0 38 32 Sample With condition 70 diff mean(x) n 38 Without condition 32 Ha: diff |t|) = F [95% Conf. Intervall 7.349579 5.9126 6.98079 .1552544 t = degrees of freedom = 9.650421 7.7874 8.510638 3.144746 E b. What is the conclusion of your test? c. Use this output to construct a 99% confidence interval for the difference between the two means. 68 Ha: diff0 Pr(Tt) = 0.0155
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
The following is a sample of patient's calcium tests in mg/dL. 9.3 8.8 10.1 8.9 9.4 9.8 10.0 9.9 11.2 12.1 Find each of the following: The mean: The median: The third quartile: The standard deviation: The variance:
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"
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
14.
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.)
An article in a Journal reports that 34% of American fathers take no responsibility for childcare. A researcher claims that the figure is higher for fathers in the town of Littleton. A random sample of 234 fathers from Littleton yielded 96 who did not help with childcare. How many populations? 01 What is the parameter? O Mean O Proportion O Difference between Means O Variance Standard Deviation
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
Refer to the below table. Using an alpha = 0.05, test the claim that IQ scores are the same for children in three different blood lead level groups: low lead level, medium lead level, and high lead level). One-Way Analysis of Variance Summary Table for IQ Measurements for Children among Three Blood Lead Level Groups: Low Lead Level, Medium Lead Level, and High Lead Level. Source df SS MS F p Between-group (treatment) 2 469.1827 2677.864 2.30 0.104 Within-group (error) 118 203.6918 11745.05 Total 120

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