64% of students will apply for graduate school. from a sample of 100 students, 52 stating that they will apply for grad school. for alpha=.01 you ___ hypothesis

Answer:

you accept the alternative hypothesis

Step-by-step explanation:

The null and alternative hypothesis can be computed as follows:

[tex]H_o: p = 0.64 \\ \\ H_1: p \le 0.64[/tex]

The sample proportion [tex]\hat p = \dfrac{x}{n}[/tex]

[tex]\hat p = \dfrac{52}{100}[/tex]

[tex]\hat p = 0.52[/tex]

The test statistics can be computed as:

[tex]Z = \dfrac{\hat p -p_o}{ \sqrt{ \dfrac{p_o \times (1-p_o)}{ n} }}}[/tex]

[tex]Z = \dfrac{0.52 -0.64}{ \sqrt{ \dfrac{0.64 \times (1-0.64)}{ 100} }}}[/tex]

Z = -2.5

The P-value = 2( Z< -2.5)

From the tables;

The P-value = 2 (0.0062)

The P-value = 0.0124

The significance level = 0.01

Since the P-value is > significance level; we fail to reject the null hypothesis.

Conclusion: we accept the alternative hypothesis

Using the z-distribution, it is found that since the test statistic is less than the critical value for the left-tailed test, it is found that there is enough evidence to conclude that the percentage is less than 64%.

At the null hypothesis, it is tested if the proportion is of 64%, that is:

[tex]H_0: p = 0.64[/tex]

At the alternative hypothesis, it is tested if the proportion is less than 64%, that is:

[tex]H_1: p < 0.64[/tex]

The test statistic is given by:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

In which:

For this problem, the parameters are: [tex]n = 100, \overline{p} = \frac{52}{100} = 0.52, p = 0.64[/tex]

Hence:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

[tex]z = \frac{0.52 - 0.64}{\sqrt{\frac{0.64(0.36)}{100}}}[/tex]

[tex]z = -2.5[/tex]

The critical value for a left-tailed test, as we are testing if the proportion is less than a value, using the z-distribution with a significance level of 0.01, is of [tex]z^{\ast} = -2.33[/tex]

Since the test statistic is less than the critical value for the left-tailed test, it is found that there is enough evidence to conclude that the percentage is less than 64%.

For more on the use of the z-distribution for hypothesis tests, you can check https://brainly.com/question/25584945

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