7.1-1. A random sample of size 16 from the normal distri- bution N(u. 25) yielded = 73.8. Find a 95% confidence interval for u.
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- 7.. need help pleaseUse the t-distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distributions are relatively normal. Test Ho : HA = Ha VS H. : HA # Hg using the fact that Group A has 8 cases with a mean of 125 and a standard deviation of 18 while Group B has 15 cases with a mean of 118 and a standard deviation of 14. (a) Give the test statistic and thep-value. Round your answer for the test statistic to two decimal places and your answer for the p- value to three decimal places. test statistic = p-value = eTextbook and Media (b) What is the conclusion of the test? Test at a 10% level. O Reject Ho. O Do not reject Ho. eTextbook and MediaQ2: Let x1,X2, . , Xn and y1, y2, ..., Ym represent two independent random samples from the respective normal distributions N(H1,07) and N (H2, 03). It is given that of = 30, but ožis unknown. Then 1. A random variable that can be used to find a 95% confidence interval for - is (x- y) - (H1 - H2) А. ns3 + ms n+m. 3(n+m-2)"tm, nm O A (x – y) – (41 – H2) В. ns? + ms; n + m. (n + m – - 2) пт (x – y) – (41 - H2) С. ns3 + mS;_n+3 3(n+ m-2) ("+3m nm Ос
- A sample of n 4 scores is selected from a population with an unknown mean. The sample has a mean of M= 40 and a variance of s² = 16. Which of the following is the correct 90% confidence interval for u? %3D O H = 40 + 1.638(4) O H = 40 + 2.353(4) OH= 40 + 1.638(2) Ομ= 40+ 2.353(2)Suppose a marketing company randomly surveyed 404 households and found that in 214 of them, the woman made the majority of the purchasing decisions. Construct a 90% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions.p'=α2=zα2=Margin of Error: E=We are 90% confident that the proportion of households in the population where women make the majority of purchasing decisions is between___ and ___.Let X1, X2, X3, ..., X, be a random sample from a distribution with known variance Var(X,) = o², and unknown mean EX, = 0. Find a (1 – a) confidence interval for 0. Assume that n is large.
- Independent samples of size n1 = 25 and n2 = 36 are taken from two normal populations with knownstandard deviations of σ1 = 5.5 and σ2 = 4.2. e sample means are x¯1 = 13.6 and x¯2 = 19.2. Find a95% confidence interval for µ1 − µ2.Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (Η1 = Η2), so that the standard error of the difference between means is obtained by pooling the sample variances. 1) A paint manufacturer wanted to compare the drying times of two different types of paint.Independent simple random samples of 11 cans of type A and 9 cans of type B were selected andapplied to similar surfaces. The drying times, in hours, were recorded. The summary statistics areas follows.Type A Type Bx1 = 71.5 hr x2= 68.5 hrs1 = 3.4 hr s2 = 3.6 hrn1 = 11 n2 = 9a). Construct a 99% confidence interval for µ1- µ2, the difference between the mean drying timefor paint type A and the mean drying time for paint type B b). Run a hypt. test to determine the results*Alt is not equal*Make sure to select pooled…