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:
![$ Z = \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }} $](https://tex.z-dn.net/?f=%24%20Z%20%3D%20%20%5Cfrac%7B%5Chat%7Bp%7D%20-%20p%7D%7B%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%20%7D%7D%20%20%24)
![$ Z = \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }} $](https://tex.z-dn.net/?f=%24%20Z%20%3D%20%20%5Cfrac%7B0.31%20-%200.40%7D%7B%20%5Csqrt%7B%5Cfrac%7B0.40%281-0.40%29%7D%7B100%7D%20%7D%7D%20%20%24)
![$ Z = \frac{- 0.09}{ 0.048989 } $](https://tex.z-dn.net/?f=%24%20Z%20%3D%20%20%5Cfrac%7B-%200.09%7D%7B%200.048989%20%7D%20%20%24)
![Z = - 1.84](https://tex.z-dn.net/?f=Z%20%3D%20-%201.84)
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.