A staffing agency for in-home nurses and support staff places necessary personnel at locations on a daily basis. Each placed nur
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Answer:
We failed to reject H₀
Z > -1.645
-1.84 > -1.645
We failed to reject H₀
p > α
0.03 > 0.01
We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.
Step-by-step explanation:
Set up hypotheses:
Null hypotheses = H₀: p = 0.40
Alternate hypotheses = H₁: p < 0.40
Determine the level of significance and Z-score:
Given level of significance = 1% = 0.01
Since it is a lower tailed test,
Z-score = -2.33 (lower tailed)
Determine type of test:
Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.
Select the test statistic:
Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.
Set up decision rule:
Since it is a lower tailed test, using a Z statistic at a significance level of 1%
We Reject H₀ if Z < -1.645
We Reject H₀ if p ≤ α
Compute the test statistic:
From the z-table, the p-value corresponding to the test statistic -1.84 is
p = 0.03288
Conclusion:
We failed to reject H₀
Z > -1.645
-1.84 > -1.645
We failed to reject H₀
p > α
0.03 > 0.01
We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.