For an inference about the difference between two population means #1-2, we know that if we have three assumptions: n₁ < 30 or n₂ < 30, normal populations and unknown population variance o2 = 0² = 0², then we will use a t-test statistic with pooled sample variance 7₁-1 n₁ + n₂ + 7₂-1 n₁+n₂-2
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- Why would manufacturers and businesses be interested in construeting a confidence interval for the population variance? Would they want large or small variances?We are running a hypothesis test to see if blood pressure of African zebras differs significantly from Arabian horses. We assume blood pressure in both populations is normally distributed and both populations have the same population variance in blood pressure which is unknown to us, so we use the sample variances to estimate the standard error for our calculations. We have a sample of size 9 for the African zebras and a sample of size 16 for the Arabian horses. How many degrees of freedom does our test statistic have?If the variance in the weights of one sample of 12 oz. cookies is 3 oz. and the variance in a second sample, made in another factory, is only 1 oz. is there a real difference in variances between the two? (Estimate at 95% probability level, using 2-tail test). Use the appropriate test statistic to determine this, and assume there are 9 degrees of freedom for each sample. Yes no maybe none of the above.
- Suppose that you want to test the mean of two populations whose variance is known. Your sample size is large and yo do not have any information about α significance level. Explain how you made this decision. If your test is right tail and test statistics value is 3.12 then what would be the P-value and your decision accordingly?We are running a hypothesis test to see if weight of African zebras differs significantly from that of Arabian horses. We assume weight in both populations is normally distributed and both populations have the same population variance in weight which is unknown to us, so we use the sample variances to estimate the standard error for our calculations. We have a sample size of 9 for the African zebras with mean 650 and a sample of size 16 for the Arabian horses with mean 658. If the pooled variance is 36, then what is the value of the standardized test statisitic for our data?Two different types of catalysts are used in a chemical process. We want to conduct a hypothesis test to see if there is a difference in the yields at 5% significance level. Assume both populations are normaly distributed. To this end, we collect a simple random sample from each process. Assume equal variances. = = n-11, n=13,x,-90, x 91.4, 8, 4, 8=4.5
- Gestation period is the length of pregnancy, or to be more precise, the interval between fertilization and birth. In Syrian hamsters, the average gestation period is 16 days. Suppose you have a sample of 31 Syrian hamsters who were exposed to high levels of the hormone progesterone when they were pups, and who have an average gestation length of 17.1 days and a sample variance of 26.0 days. You want to test the hypothesis that Syrian hamsters who were exposed to high levels of the hormone progesterone when they were pups have a different gestation length than all Syrian hamsters. Calculate the t statistic. To do this, you first need to calculate the estimated standard error. The estimated standard error is sMM= a. 31, b. 0.5232, c. 0.7328, d. 0.9158. The t statistic is- a. 1.50, b. 2.10, c. 2.63, d. 1.20 Now suppose you have a larger sample size n = 95. Calculate the estimated standard error and the t statistic for this sample with the same sample average and the same…A study aims to compare the social media usage rates of men and women. Nine men and 9 women are drawn from the general population and the average time (in minutes) they spent on social media each day was compared (men’s mean = 40, sum of squares = 153; women’s mean = 48, sum of squares = 216). What is the (a) estimated population variance for each random sample, and (b) pooled variance? Estimated population variances = 19.13 and 27.00; Pooled variance = 23.07 Estimated population variances = 19.13 and 27.00; Pooled variance = 20.50 Estimated population variances = 17.00 and 24.00; Pooled variance = 20.50 Estimated population variances = 17.00 and 24.00; Pooled variance = 23.07In a comparison between the mean waiting time of the patients in the hospitals of Giza and Alexandria governments the minister of Health randomly selected 60 hospitals from Giza government and their mean waiting time was 30 minutes and the previous reports showed that the variance was 10 minutes. Also a sample consist of 50 hospitals from Alexandria governments was selected and their mean waiting time of this sample was 15 minutes and from previous reports the variance was 5 minutes. 1-Construct the 95% C.I for the difference between the average waiting time in Giza & Alex2- Without any computations, discuss fully and precisely what will happen to the results you obtained in part (1) if the confidence level changed to be 99% instead of 95%. Give the reason for your answer
- If, in an analysis of variance, you find that MSTR is greater than MSE, why can you not immediately reject the null hypothesis without determining the F ratio and its distribution? Explain.A sample with SS=30 and a variance of s2 =10 has n=4 scores. True or False?ANOVA is used for comparing across 3 or more groups.In ANOVA, the null hypothesis is always: at least one variance is different. at least one mean is different. all means are equal. all variances are equal