Mathematics, 23.07.2020 03:01 maddieg5585

A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 290 people over the age of 55, 68 dream in black and white and among 288 people under the age of 25, 19 dream in black and white. Use a 0.05 significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion of those underIdentify the test statistic?Identify the p value?Test the claim by constructing an appropriate confidence level?What is the conclusion base on the hypothesis test?What is the conclusion base on the confidence level?

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Answer from: rishamoore

he proportion of people over 55 who dream in black and white is greater than the proportion of those under.

The proportion of people over 55 who dream in black and white lies in the range (0.112, 0.226).

Step-by-step explanation:

In this case we need to determine if the proportion of people over 55 who dream in black and white is greater than the proportion of those under.

The hypothesis can be defined as follows:

H₀: The proportion of people over 55 who dream in black and white is not greater than the proportion of those under, i.e. p₁ - p₂ ≤ 0.

Hₐ: The proportion of people over 55 who dream in black and white is greater than the proportion of those under, i.e. p₁ - p₂ > 0.

The information provided is:

n₁ = 290

n₂ = 288

X₁ = 68

X₂ = 19

Compute the sample proportions and total proportions as follows:

$\hat p_{1}=\frac{X_{1}}{n_{1}}=\frac{68}{290}=0.235\\\\\hat p_{2}=\frac{X_{2}}{n_{1}}=\frac{19}{288}=0.066\\\\\hat P=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{68+19}{290+288}=0.151$

Compute the test statistic value as follows:

$z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat P(1-\hat P)[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}$

$=\frac{0.235-0.066}{\sqrt{0.151(1-0.151)[\frac{1}{290}+\frac{1}{288}]}}\\\\=5.67$

The test statistic value is 5.67.

The decision rule is:

The null hypothesis will be rejected if the p-value of the test is less than the significance level.

Compute the p-value as follows:

$p-value=P(Z5.67)\\=1-P(Z$

The p-value of the test is quite small.

The null hypothesis will be rejected at 5% significance level.

Thus, the proportion of people over 55 who dream in black and white is greater than the proportion of those under.

The significance level of the test is 5%.

Then the confidence level will be:

Confidence level = 100% - Significance level

= 100% - 5%

= 95%

Compute the 95% confidence interval for the difference between proportions as follows:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p_{1}(1-\hat p{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p{2})}{n_{2}}}$

The critical value of z for 95% confidence level is z = 1.96.

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p_{1}(1-\hat p{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p{2})}{n_{2}}}$

$=(0.235-0.066)\pm1.96\cdot\sqrt{\frac{0.235(1-0.235)}{290}+\frac{0.066(1-0.066)}{288}}\\\\=0.169\pm 0.057\\\\=(0.112, 0.226)$

The null hypothesis would be rejected if the null value, i.e. (p₁ - p₂) ≤ 0 is not contained in the interval.

The 95% confidence interval consist of values greater than 0.

Thus, the null hypothesis will be rejected.

Concluding that the proportion of people over 55 who dream in black and white lies in the range (0.112, 0.226).

Answer from: Quest

c: lkllmn.

hope it's you

What additional piece of information do you need to prove that lkn~knm by the sas similarity theorem

Answer from: Quest

the answer is 2/54

step-by-step explanation:

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