A study was conducted to determine the proportion of people who dream in black a

Question

A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 310 people over the age of 55, 74 dream in black and white, and among 306 people under the age of 25, 16 dream in black and white. Use a 0.01 significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25. Complete parts (a) through (c) below.

b. Test the claim by constructing an appropriate confidence interval. The 98%

confidence interval is blank <(p1-p2) < blank

(Round to three decimal places as needed.) I need the blanks rounded to 3 decimal places!

Explanation / Answer

P1 = 74/310 = 0.24

P2 = 16/306 = 0.05

H0: P1 = P2

H1: P1 > P2

Pooled sample proportion (P) = (P1 * n1 + P2 * n2)/(n1 + n2)

= (0.24 * 310 + 0.05 * 306)/(310 + 306) = 0.15

SE = sqrt (P * (1 - P) * (1/310 + 1/306))

= sqrt (0.15 * 0.85 * (1/310 + 1/306))

= 0.029

The test statistic Z = (P1 - P2)/SE

= (0.24 - 0.05)/0.029

= 6.55

At 98% confidence interval the critical value is 2.33

As the test statistic value is greater than the critical value so the null hypothesis is rejected.

So the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25 is accepted.

Th 98% confidence interval is

P1 - P2 +/- z* * SE

= (0.24 - 0.05) +/- 2.33 * 0.029

= 0.19 +/- 0.068

= 0.122, 0.258

As the confidence interval doesn't contain the hypothized value 0, SO the null hypothesis is rejected.

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