Answer to Question #350555 in Statistics and Probability for dexcer

Question

The principal of a school claims that 30% of the grade 11 students eat lunch

in school. A survey among 500 grade 11 students revealed that 150 of them

stayed in school during lunch. Use 95% confidence to conduct a test of

proportions.

1.645

z = 3.294

13

Step 1:Statistical

Hypotheses and

Direction of Test

H0 :

H1 :

Statistical Test :

Direction of Test :

Step 2: Level of Significance

and Critical Value

α :

zCV :

Step 3: Test Statistic zTV :

Step 4: Normal curve

Step 5: Findings and

Decision

Step 6: Interpretation

Step 7: Conclusion

Answer 1

Step 1

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p=0.3$

$H_a:p\not=0.3$

This corresponds to a two-tailed test, for which a z-test for one population proportion will be used.

Step 2

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z: |z| > 1.96\}.$

Step 3

The z-statistic is computed as follows:

$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{150/500-0.3}{\sqrt{\dfrac{0.3(1-0.3)}{500}}}=0$

Step 4

Since it is observed that $|z| = 0 \le z_c = 1.96,$ it is then concluded that the null hypothesis is not rejected.

Step 5

Using the P-value approach: The p-value is $p=2P(Z<0)=1,$ and since $p = 1 \ge 0.05,$ it is concluded that the null hypothesis is not rejected.

Step 6

It is concluded that the null hypothesis Ho is not rejected.

Step 7

Therefore, there is not enough evidence to claim that the population proportion $p$ is different than $0.3,$ at the $\alpha = 0.05$ significance level.