Question

According to a 2018 survey by a research group, 30% of adults typically run the water...

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.

  1. Identify the claim and state the H0 and H1.
  2. Find the critical value.
  3. Calculate the test statistic.
  4. Make a decision to reject or fail to reject the H0.

Interpret the decision in the context of the original claim

Homework Answers

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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