A local elementary school claims that 59% of its students bring their own lunch on a daily
basis. The principal surveyed 100 students and found that only 48% bring their own lunch.
What does the test statistic calculation look like in this situation for an appropriate
hypothesis test? (4 points)

Answer:

[tex]z=\frac{0.48-0.59}{\sqrt{\frac{0.59(0.41)}{100}}}[/tex]

Step-by-step explanation:

The z score in terms of the null hypothesis and surveyed statistic is in the form [tex]z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}[/tex], (similar to [tex]z=\frac{x-\mu}{\sigma}[/tex]) where [tex]p_0[/tex] is the population parameter (in this case, 0.59) and [tex]\hat{p}[/tex] is the statistic (in this case, 0.48). Finally, [tex]n[/tex] is the sample size, 100. Substitute in your values, and you get the above answer.

The test statistic calculation look like in this situation for an appropriate hypothesis test is negative.

A local elementary school claims that 59% of its students bring their own lunch on a daily basis.

The z score in terms of the null hypothesis and surveyed statistic is in the form,

[tex]\frac{P-P_0}{\sqrt{P_0(1-P_0)/n} }[/tex]

where [tex]p_0[/tex] is the population parameter (in this case, 0.59) and P is the statistic (in this case, 0.48)

Finally,  is n the sample size, 100.

Substitute in your values, and you get the above answer,

[tex]Z=\frac{0.40-0.59}{\sqrt{\frac{0.59(0.41}{100} } }\\ \\=-3.8630[/tex]

The test statistic calculation look like in this situation for an appropriate hypothesis test is negative.

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