Respuesta :
Answer:
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:
[tex]\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[/tex]
Compute the test statistic value as follows:
[tex]z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat P(1-\hat P)[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}[/tex]
[tex]=\frac{0.235-0.066}{\sqrt{0.151(1-0.151)[\frac{1}{290}+\frac{1}{288}]}}\\\\=5.67[/tex]
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:
[tex]p-value=P(Z>5.67)\\=1-P(Z<5.67)\\=1-\approx1\\=0[/tex]
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:
[tex]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}}}[/tex]
The critical value of z for 95% confidence level is z = 1.96.
[tex]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}}}[/tex]
[tex]=(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)[/tex]
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).
