Let's tackle this problem step-by-step to find the P-value and determine whether to reject or fail to reject the null hypothesis, given the test statistic [tex]\( z = 1.52 \)[/tex] and a significance level of [tex]\( \alpha = 0.05 \)[/tex].
1. Identify the Hypothesis:
- Null Hypothesis ( [tex]\( H_0 \)[/tex] ): [tex]\( p = \frac{4}{5} \)[/tex]
- Alternative Hypothesis ( [tex]\( H_1 \)[/tex] ): [tex]\( p \neq \frac{4}{5} \)[/tex]
This is a two-tailed test.
2. Given Information:
- Test statistic [tex]\( z = 1.52 \)[/tex]
- Significance level [tex]\( \alpha = 0.05 \)[/tex]
3. Calculate the P-value:
- For a two-tailed test, the P-value is calculated as:
[tex]\[
P = 2 \times \left(1 - \text{CDF}(z)\right)
\][/tex]
where CDF is the cumulative distribution function of the standard normal distribution.
- From statistical tables or computations, for a [tex]\( z = 1.52 \)[/tex], the cumulative distribution function [tex]\( \text{CDF}(1.52) \approx 0.9357 \)[/tex].
- Therefore,
[tex]\[
P = 2 \times (1 - 0.9357) = 2 \times 0.0643 = 0.1286
\][/tex]
4. Compare the P-value with [tex]\( \alpha \)[/tex]:
- The P-value is [tex]\( 0.1286 \)[/tex].
- Since [tex]\( 0.1286 > 0.05 \)[/tex], we do not have enough evidence to reject the null hypothesis at the [tex]\( \alpha = 0.05 \)[/tex] significance level.
5. Conclusion:
- Given that the P-value [tex]\( 0.1286 \)[/tex] is greater than the significance level [tex]\( \alpha = 0.05 \)[/tex], we fail to reject the null hypothesis.
6. Matching with the Choices Given:
- The correct option from the choices is:
[tex]\[
(0.1286, \text{"fail to reject the null hypothesis"})
\][/tex]
Thus, the P-value is [tex]\( 0.1286 \)[/tex] and the conclusion is to fail to reject the null hypothesis.
