A company that makes shampoo wants to test whether the average amount of shampoo per bottle is 16 ounces. The standard deviation is known to be 0.20 ounces. Assuming that the hypothesis test is to be performed using 0.10 level of significance and a random sample of n = 64 bottles, how large could the sample mean be before they would reject the null hypothesis?

Question

BlackTom AI · Accepted Answer

We start by identifying the test setup. The company tests whether the average amount of shampoo per bottle mu equals 16 ounces under a known standard deviation sigma = 0.20, with n = 64 and alpha = 0.10 for a two-sided test.

1) Compute the standard error of the mean: SE = sigma / sqrt(n) = 0.20 / sqrt(64) = 0.20 / 8 = 0.025 ounces.

2) For a two-sided test at alpha = 0.10, the critical z-value is z_{0.95} ≈ 1.6449 (since you split alpha/2 = 0.05 in each tail and look up the 95th percentile).

3) The rejection region is |Xbar - mu0| > z * SE. Here mu0 = 16, so the threshold is |Xbar - 16| > 1.6449 * 0.025.

4) Multiply: 1.6449 * 0.025 ≈ 0.0411225 ounces.

5) Therefore, the sample mean would need to exceed 16 by more than about 0.0411 ounces to reject H0. The cutoff value (the largest mean that would not lead to rejection on the high side) is 16 + 0.0411 ≈ 16.041 ounces.

Option analyses:
- 16.2 ounces: This is far above the cutoff of 16.041; a mean this large would lead to rejection, not the maximum non-rejecting value.
- 15.8 ounces: This is below the null mean by 0.2 ounces, which would certainly lie in the non-rejection region for the left tail, but it does not represent the upper threshold before rejection on the high side.
- 16.049 ounces: This is very close to the cutoff but slightly larger than 16.041; since the exact threshold is about 16.041, 16.049 would exceed the non-rejection boundary and would lead to rejection.
- 16.041 ounces: This matches the calculated boundary where |Xbar - 16| = 0.0411, i.e., the largest mean that would not cause rejection on the high side; it aligns with the two-sided 0.10 significance level boundary.

In summary, the cutoff value derived from SE and the z-critical corresponds to a maximum non-rejecting mean of approximately 16.041 ounces, making 16.041 ounces the correct choice among the options. The other values either lie inside or outside the rejection region in ways that do not match the calculated boundary.

OSC 2030-009 Fall 2025 SUBMIT: Graded Homework Module 8 (Chap 9)

查看解析

登录即可查看完整答案

类似问题

Which one is true?

For an upper-tail hypotheses about a population mean with σ known and α=5%, which one is the rejection region?

For a given sample, the P-value for upper tail test is x, and the P-value for two-tailed test is y, then

For a two-tailed hypothesis test with a test statistic value of z = 2.09, the p-value is _____.

For a sample size of 30, changing from using the t distribution to using the standard normal distribution in a hypothesis test _____.

Exhibit 9-5 n = 16 H0: μ ≥ 80 = 75.607 Ha: μ < 80 σ = 8.6 Assume the population is normally distributed. Refer to Exhibit 9-5. The p-value is equal to _____.

Exhibit 9-1 Refer to Exhibit 9-1. If the test is done at a .01 level of significance, the null hypothesis should _____.

What conclusion for a test would correctly detect a difference between the actual parameter and the hypothesized value for the parameter? c08.p089.q0008

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单