Test the hypothesis using the P-value approach.

Question

Test the hypothesis using the P-value approach.

Falak Kinney 2021-05-27 Answered

Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test.
$H_{0} : P = 0.52$ versus $H_{1} : p < 0.52$
$N = 150, X = 72, α = 0.1$
Is $N P_{0} (1 - P_{0})$ greater than or equal to 10?
Use technology to find the P-Value.

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Expert Answer

pattererX

Answered 2021-05-28 Author has 95 answers

Null hypothesis, $H_{0} : P = 0.52$
Alternative hypothesis, $H_{1} : P < 0.52$
Under $H_{0}$ , the population proportion is $P_{0} = 0.52$
Now consider,
$N P_{0} (1 - P_{0}) = 150 \times 0.48 \times (1 - 0.48)$
$= 37.44$
$> 10$
Since $N P_{0} (1 - P_{0})$ is greater than or equals to 10, use normal approximation to test the true proportion.
Given information:
Number of trials, $N = 150$
Number of sucesses in 150 cases, $X = 72$
Sample proportion,
$p = \frac{X}{N}$
$= \frac{72}{150}$
$= 0.48$
Compute test statistic.
$z = \frac{p - P_{0}}{\sqrt{\frac{P_{0} (1 - P_{0})}{N}}}$
$= \frac{0.48 - 0.52}{\sqrt{\frac{(0.52) (1 - 0.52)}{N}}}$
$= - 0.98058$
$= - 0.98$
Alternative hypothesis has symbol "<", so the test is based on left-tailed.
Using Excel function "=NORMSTD(z)"
$P - v a l u e = P (Z \leq z)$
$= P (Z \leq - 0.98)$
$= N O R M S D I S T (- 0.98)$
$= 0.1635$
When testing the hypothesis $H_{0} : P = 0.52$ versus $H_{1} : P < 0.52$ , using the sample information $N = 150$ and $X = 72$ , the P-value of test is 0.1635

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