Test the claim that the average weight of a new SUV is 1,940 kg, if a sample of 309vehicle weights results in a sample mean of 1,931 kg, with a standard deviation of 61.7kg. Use a 5% level of significance.Standard Normal Distribution Tablea. Calculate the test statistic.0.00Round to two decimal places if necessaryb. Determine the critical value(s) for the hypothesis test.Round to two decimal places if necessaryc. Conclude whether to reject the null hypothesis or not based on the test statistic.O RejectO Fail to Reject

Given,sample size(n)=309sample mean(x¯)=1,931standard deviation(s)=61.7α=0.05and degrees of…

Answered: Test the claim that the average weight…

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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Moviegoers The average "moviegoer" sees 8.5 movies a year. A moviegoer is defined as a person who sees at least one movie in a theater in a 12-month period. A random sample of 42 moviegoers from a large university revealed that the average number of movies seen per person was 10.2. The population standard deviation is 3.2 movies. At the 0.01 level of significance, can it be concluded that this represents a difference from the national average? Part 1 of 5 State the hypotheses and identify the claim with the correct hypothesis. Hotu = 85 Part: 1/5 not claim H₁ 8.5 claim This hypothesis test is a two-tailed ▼test. Part 2 of 5 ▼ ▼ Find the critical value(s). Round the answer to at least two decimal places. If there is more than one critical value, separate them with commas. Critical value(s): X 0.0.... + 8分民图〗b. E
The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces.Refer to Exhibit 6-5. What is the random variable in this experiment? a. 8 ounces b. Weight of items produced by a machine c. Normal distribution d. 2 ounces
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
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The mean ACT score for 43 male high school students is 21.1 and the standard deviation is 5.0. The mean ACT score for 56 female high school students is 20.9 and the standard deviation is 4.7. At the 1% significance level, can you reject the claim that male and female high school students have equal ACT score averages? Be sure to use the p-value to make your conclusion.
Americans are about 1 inch taller but nearly 25 pounds heavier than in 1960. The average BMI increased from approximately 25 to 28 in 2002. Boston is considered one of americas healthiest cities. Is the weight gain since 1960 similar in boston? Sample of n=25 and mean increase of 17 with standard deviation of 8.6. Is boston statistically significantly different in terms of weight gain since 1960? Run the appropriate test at 5%level of significance. A.state the null and the research hypotheses, b. Idebtify critical values of z 9r t used to define the decision rule, c. Calculate the test statistic, d. Report the p-value, e. Report your findings
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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.)
When 40 people used the Weight Watchers diet for one year, their mean weight loss was 3.0 pounds and the standard deviation was 4.9 pounds. Use a 0.01 significance level to test the claim that the mean weight loss is greater than zero. D. Calculate the P-value E. State your decision about the hypotheses F Write a final, concluding statement that addresses the claim: There is _______________ Evidence to _________________ the claim that the mean weight loss is greater than 0 punds.

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