Question 4 (18 marks) Putting a baby to sleep can be a nightmare for parents. A

Question

Question 4 (18 marks) Putting a baby to sleep can be a nightmare for parents. A company developed a program called "Sleeping Well" and claimed that the program was effective to more than half of the babies. One hundred and twenty families participated in the trial. Of these, 65 families found the program effective. (a) Calculate the sample proportion of families who found the program effective. 1 mark (b) Using the sample proportion, calculate the number of families needed to construct a 95% 3 marks (c) By means of a hypothesis test, determine whether or not the company's claim was sup- confidence interval for a similar trial so that the margin of error is no greater than 0.03. 6 marks) The company found that "Sleeping Well" did not generate as much revenue as they expected. ported by the data. Use the 5% level of significance. After some further development, they launched "Sleeping Well 2" Another trial was conducted. This time, 200 families participated and 135 of them found the program effective. (d) Calculate the sample proportion of families who found "Sleeping Well 2" effective. [1 mark] (e) Combining the information of the two trials, calculate the pooled proportion of families [1 mark (1) Determine whether not there was an improvement in the effectiveness of "Sleeping Well (6 marks] who found "Sleeping Well" or "Sleeping Well 2" effective. 2" by conducting a hypothesis test. Use the 5% level of significance.

Explanation / Answer

(a) Sample proportion of families who found program effective = 65/120 = 0.5417

(b) Here sample size = n

margin of error = 0.03

Critical test statistic for 95% confidence interval = 1.96

so,  

0.03 = 1.96 * sqrt [0.5417 * (1 - 0.5417)/120]

n = (1.96/0.03)2 * 0.5417 * (1 - 0.5417) = 1005.63 or 1006

(c) Here standard error of proportion = sqrt(0.5 * 0.5/120) = 0.0456

test stastic

Z = (0.5417 - 0.5)/0.0456 = 0.9145

p- value = Pr(Z > 0.9145) = 1 - Pr(Z < 0.9145) = 1 - 0.8198 = 0.1802

here for 0.05 level of significance, the p - value is greater than 0.05 so here we reject the null hypothesis that sleeping well is ineffective.

(d) Here sample proportion = 135/200 = 0.675

(e) Pooled proportion p = (135 + 65)/(120 + 200) = 0.625

(f) Here now we will test that there was an improvement in effectiveness of "Sleeping Well 2" by conduction a hypothesis test.

standard error of propotion = sqrt[p * (1 - p) * (1/n1 + 1/n2)] = sqrt [0.625 * 0.375* (1/120 + 1/200)]= 0.0559

Test statistic

Z = (0.675 - 0.5417)/0.0559 = 2.3846

p - value = Pr(Z > 2.3846) = 1 - 0.9915 = 0.0085

so here p - value is less than 0.05 so we can say that there is an improvement in the effectiveness of "Sleeping well 2" compared to "Sleeping Well 1".

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