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In: Statistics and Probability

For the following problem: A) State the Null and Alternative hypotheses. B) Identify the type of...

For the following problem:

A) State the Null and Alternative hypotheses.

B) Identify the type of test (left tailed, Right Tailed, Two tailed) and state the level of significance.

C) Calculate the p-value and make a decision to either reject/not reject the null hypothesis.

D) Interpret your decision in the context of the problem.

In a random sample of 100 adult Americans who did not attend college, 37 said they believe in extraterrestrials. In a random sample of 100 adult Americans who did attend college, 47 said they believe in extraterrestrials. At the 1% level of significance, does this indicate that the proportion of people who did not attend college and believe in extraterrestrials is less than the proportion of people who did attend college and believe in extraterrestrials?

Solutions

Expert Solution

A).

p₁ : proportion of people who did not attend college and believe in extraterrestrials

p₂ : proportion of people who did attend college and believe in extraterrestrials

Claim : proportion of people who did not attend college and believe in extraterrestrials is less than the proportion of people who did attend college and believe in extraterrestrials

Null hypothesis : H_o : p₁ = p2

Alternate Hypothesis : H_a : p₁ < p₂

B).

Left tailed test

level of significance = 1%

Using pooled estimation of the proportion

$Test\; \; Statistic : Z=\frac{\widehat{p_{1}}-\widehat{p_{2}}}{\sqrt{\widehat{p_{p}}(1-\widehat{p_{p}})}\sqrt{\frac{1}{n_{1}}+ \frac{1}{n_{2}}}}$

$Pooled\; \; Sample\; \; Proportion \; :\widehat{p_{p}}=\frac{x_{1}+x_{2}}{n_{1}+n_{2}}$

Given
n₁ : Sample Size of Sample 1: Number of adult Americans of the sample who did not attend college	100
n₂ : Sample Size of Sample 2: Number of adult Americans of the sample who did attend college	100
x₁ : 37 said they believe in extraterrestrials from sample 1	37
x₂ :47 said they believe in extraterrestrials from sample 1	47
$\widehat{p}_{1}$ : Sample Proportion of of people who did not attend college and believe in extraterrestrials	=37/100 = 0.37
$\widehat{p}_{2}$ : Sample Proportion of of people who did attend college and believe in extraterrestrials	=47/100 = 0.47
$Level \;of \;Significance\; :\; \alpha$	1%

$Pooled\; Sample\; Proportion\; :\; \widehat{p_{p}}=\frac{x_{1}+x_{2}}{n_{1}+n_{2}}$

$\widehat{p_{p}}=\frac{x_{1}+x_{2}}{n_{1}+n_{2}} =\frac{37+47}{100+100}=\frac{84}{200}=0.42$

$Test\; \; Statistic : Z=\frac{\widehat{p_{1}}-\widehat{p_{2}}}{\sqrt{\widehat{p_{p}}(1-\widehat{p_{p}})}\sqrt{\frac{1}{n_{1}}+ \frac{1}{n_{2}}}}$

$Test\; \; Statistic : Z=\frac{0.37-0.47}{\sqrt{0.42(1-0.42)}\sqrt{\frac{1}{100}+ \frac{1}{100}}}=\frac{-0.1}{0.0698}=-1.4327$

for left tailed test :

$P-Value = P(Z<Z_{stat}) = P(Z<-1.4327)=0.076$

As P-Value i.e. is greater than Level of significance i.e (P-value:0.076 > 0.01:Level of significance); Fail to Reject Null Hypothesis

D)There is not sufficient evidence to conclude that that the proportion of people who did not attend college and believe in extraterrestrials is less than the proportion of people who did attend college and believe in extraterrestrials

Alternatively,

C) If the proportions are estimated separately.

$Test\; \; Statistic : Z=\frac{\widehat{p_{1}}-\widehat{p_{2}}}{\sqrt{\frac{\widehat{p_{1}}(1-\widehat{p_{1}})}{n_{1}}+ \frac{\widehat{p_{2}}(1-\widehat{p_{2}})}{n_{2}}}}$

$Test\; \; Statistic : Z=\frac{0.37-0.47}{\sqrt{\frac{0.37(1-0.37)}{100}+ \frac{0.47(1-0.47)}{100}}}=\frac{-0.1} {0.0694}=-1.4401$

For left tailed test :

$P-Value = P(Z<Z_{stat}) = P(Z<-1.4401)=0.0749$

As P-Value i.e. is greater than Level of significance i.e (P-value:0.0749 > 0.01:Level of significance); Fail to Reject Null Hypothesis

orchestra answered 2 years ago

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