Question

According to a 2018 survey by a research group, 30% of adults typically run the water for a period of 6 to 10 minutes while taking the shower). Suppose that in a recent survey of 400 adults, 104 stated that they typically run the water for a period of 6 to 10 minutes when they take a shower. At the 5% significance level, can you conclude that the proportion of all adults run the water for a period of 6 to 10 minutes when they take a shower is less than 0.30? Answer the following questions.

- Identify the claim and state the H
_{0}and H_{1}. - Find the critical value.
- Calculate the test statistic.
- Make a decision to reject or fail to reject the
H
_{0}.

Interpret the decision in the context of the original claim

Answer #1

Ans. Let

Population proportion of adults who typically run the water for a period of 6 to 10 minutes, P = 30% or 0.3

Sample proportion of adults who typically run the water for a period of 6 to 10 minutes, p = 104/400 = 0.26

Population standard error, se = [P*(1-P)/Sample Size]^0.5 = 0.0229128

a) Hypotheses,

H0: P > = 0.30, against

H1: P < 0.30 (Lower tailed test)

b) At 5% level of significance for a lower tailed test, z critical = -1.645

c) Test statistic, z because sample size is very large,

so, z statistic = (p - P)/se = (0.26 - 0.30)/0.0229128 = -1.74574

d) As |z statistic| > |z critical|, so, we reject the null hypothesis and hence, the proportion of all adults run the water for a period of 6 to 10 minutes when they take a shower is less than 0.30

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