Test the null hypothesis that the mean weight of all Sain Bernard dogs is 200 pounds (use the signigicance level a=0.05). State the conclusion in terms of the problem and calculate the P-value for this test.
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A simple random sample of 36 Saint Bernard dogs weight yield a sample
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- Statistics QuestionA psychologist is studying smokers' self-images, which she measures by the self-image (SI) score from a personality inventory. For adults in the U.S., the mean SI score from this inventory is about 150.The psychologist gathers a random sample of 15 SI scores of smokers and finds that their mean is 126 and their standard deviation is 37. Assume that the population of SI scores of smokers is normally distributed with mean μ. Based on the sample, can the psychologist conclude that μ is different from 150? Use the 0.05 level of significance. Perform a two-tailed test. Then fill in the table below.Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. The null hypothesis: H0: The alternative hypothesis: H1: The type of test statistic: (Choose one) Z t Chisquare F The value of the test statistic:(Round to at least three decimal places.) The p-value:(Round to at least…Find the standard error of the sample mean such that we have 70 random samples with mean 40 and standard deviation of 16.
- When 40 people used the Weight Watchers dies for one year, their mean weight loss was 3.0 lb and the standard deviation was 4.9 lb. Use a 0.01 significance level to test the claim that the mean weight loss is greater than 0. (Show the solution) A. The null hypothesis is i. The mean weight loss is equal to 0. ii. The mean weight loss is not equal to 0. iii. The mean weight loss is greater than 0. iv. The mean weight loss is less than 0. B. The computed t-test statistic is i. 4.468 ii. 4.977 iii. 3.872 iv. 3.391 C. The critical/tabular t-test statistic is i. 2.426 ii. 1.684 iii. 1.685 iv. 2.423 D. Decision rule: i. Do not reject the null hypothesis. ii. Reject the null hypothesis. E. The final conclusion is i. The mean weight loss is not equal to 0 therefore Weight Watchers is effective. ii. The mean weight loss is less than 0 therefore Weight Watchers is not effective. iii. The mean weight loss is equal to 0 therefore Weight Watchers…Does it take a different amount of time for seeds to germinate if they are near rock music that is continuously playing compared to being near classical music? The 48 seeds that were exposed to rock music took an average of 28 days to germinate. The standard deviation was 13 days. The 47 seeds that were exposed to classical music took an average of 32 days to germinate. The standard deviation for these seeds was 10 days. What can be concluded at the a - 0.01 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: Select an answer Select an answers Select an answerv (please enter a decimal) Hị: Select an answer Select an answer Select an answer v (Please enter a decimal) c. The test statistic ?- (please show your answer to 3 decimal places.) d. The p-value- (Please show your answer to 4 decimal places.) e. The p-value is ? a f. Based on this, we should Select g. Thus, the final conclusion is that . answer v the null…Do the mean weights of male college freshmen and male college seniors differ? There are arguments for and against this question. The mean weight, μx of college freshmen will be compared to the mean weight, My of college seniors. The true values of Mx and My are unknown. It is recognized that the true standard deviations are x = 13 for male college freshmen and dy = 24 for male college seniors. We take a random sample of m = 431 college freshmen and a random sample of n=289 college seniors. The mean weights were x = 178 for male college freshmen and y = 185 for male college seniors. Assuming independence between the samples and assuming the weights are normally distributed we would like to estimate My - My. a) What is the standard deviation of the distribution of x?[ b) What is the standard deviation of the distribution of x-y? c) Create a 95% confidence interval for Mx - Hy? d) What is the length of the confidence interval in part c) ? e) If we let n stay at 289 but vary m, what is the…
- A random sample of 32 items from the first population showed a mean of 112 and a standard deviation of 19. A sample of 28 items for the second population showed a mean of 106 and a standard deviation of 8. Test the assumption that the population variances are equal with 0.10 significance level. Then, using a 0.05 significance level, test the hypothesis that the population means are equal. Does it take less time for seeds to germinate if they are near rock music that is continuously playing compared to being near classical music? The 43 seeds that were exposed to rock music took an average of 27 days to germinate. The standard deviation was 10 days. The 59 seeds that were exposed to classical music took an average of 34 days to germinate. The standard deviation for these seeds was 14 days. What can be concluded at the a = 0.05 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: Select an answer v Select an answer v Select an answer v (please enter a decimal) H1: Select an answer v Select an answer v Select an answer v (Please enter a decimal) c. The test statistic ?v= (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is ? va f. Based on this, we should Select an answer v the null hypothesis. g. Thus, the final conclusion is…A psychologist investigated whether there is a difference, on average, in IQ scores for twins. She randomly selected 30 sets of twins from a database and measured their IQs. The mean difference (firstborn – second-born) was 2.1 points with a standard deviation of 7.2 points. Do the data provide a convincing evidence of a difference, on average, in IQ scores for twins? Do Perform the calculations, then complete the conclusion. Conclude Because the P-value, which is , is = 0.05, we H0. We convincing evidence of , on average, in IQ scores for twins.
- A peditrician claimed that the average time for a drug to take effect is 15 minutes with standard deviation of eight minutes. In sample of 49 trials the average time was 18 minutes. Test the claim of a pediatrician that the average time is not equal to 15 minutes. Assume the the average time for a drug to take effect is approximately normal distributedThe mean SAT score in mathematics, u, is 596. The standard deviation of these scores is 47. A special preparation course clalms that its graduates will score higher, on average, than the mean score 596. A random sample of 90 students completed the course, and their mean SAT score in mathematics was 608. At the 0.05 level of significance, can we condude that the preparation course does what it claims? Assume that the standard deviation of the scores of course graduates is also 47. Perform a one-talled test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H, and the alternative hypothesis H,. H, :0 H, :0 (b) Determine the type of test statistic to use. OSO (Choose one) ▼ (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the critical value. (Round to three or more decimal…A recent study stated that if a person chewed gum, the average number of sticks of gum he or she chewed daily was 8. To test the claim, the researcher selected a random sample of 36 gum chewers and found the mean number of sticks of gum chewed per day was 8. The standard deviation of population is 1. at a = 0. 05, is the number of sticks of gum a person chews per day actually greater than 8. There is not enough evidence to support the claim that average is greater than 8. There is enough evidence to support the claim that average is greater than 8. There is enough evidence to support the claim that average is smaller than 8. There is not enough evidence to support the claim that average is greater than 8.