Question

CLO'S 6: Setup and sest hy Exercisest In 2011, it has been found that the daily production workers for of this company is equal to $112. A sample of 90 production workers company revealed a sample mean of S117 per day. Assuming the lude that there is an 0.05, can we conclude that tpopulation standard deviation $11.2 At α increasein the daily workers earnings? Exercises#l mean A real estate research company tracks the cost of apartment rental. The apartment rate is $895 per month. Assume that based on historical population standard deviation is $225. In a recent study of apartment rental costs, a sample of 180 apartments provides a sample mean of $885 per month At the significance level α 0.01, can we conclude that decrease occurred in the mean apartment rental cost? surveys, the a) Formulate the null and alternative hypothesis b) At a significance level of a -001, give your conclusion. Exercises The manager at Air Express thinks that weights of packages shipped recently are less than in the past. Records show that in the past packages have had a mean weight of Ho 16.7 KG and a standard deviation of6.5 KG, a random sample of last month's shipping records yielded a mean weight of X- 14.6 KG for 64 packages. Is this sufficient evidence to reject the null hypothesis in favor of the manager's claim? a) Construct the null and alternative hypothesis. b) At significance level of a 0.01, give your conclusion. Exercises#4 It has been claimed that the mean weight of women students at a college is 55.5Kg. Professor Ali does not believe the statement and sets out to show the mean weight is not 55.5Kg. To test the claim he collects a random sample of 200 weights from among the women students. A sample means of 54.75kg results. Is this sufficient evidence for Professor Ali to reject the statement? Use α-0.01 and σ 5.5Kg.

Answer 1

1.

Null Hypothesis H0: Mean daily workers earnings = $112
Alternative hypothesis H1: H0: Mean daily workers earnings > $112

Level of significance = 0.05

As, we know the population standard deviation, we will conduct one sample z test to conduct the hypothesis test.

Standard error of sample mean = $\sigma / \sqrt{n} = 11.2 / \sqrt{90} = 1.180584$

Test statistic, z = (Observed mean - Hypothesized mean) / standard error

= (117 - 112) / 1.180584

= 4.235

For right tail test, p-value = P(z > 4.235) = 0.00001

As, p-value is less than the significance level of 0.05, we reject null hypothesis H0 and conclude that there is significant evidence that the mean daily workers earnings is greater than $112 and thus there is is significant evidence that there is increase in mean daily workers earnings.