Question

4. In a survey sponsored by the Lindt chocolate company, 1708 women were surveyed and 85% of them said that chocolate made them happier. (a) Is there anything potentially wrong with this survey? (b) Of the 1708 women surveyed, what is the number of them who said that chocolate made them happier? (c) Use Excel to construct a 98% confidence interval estimate of the percentage of women who say that chocolate makes them happier. Insert a screenshot, write down the confidence interval and write a brief statement interpreting the result. 3 (d) Use Excel to test the claim that when asked, more than 80% of women say that chocolate makes them happier. Use a 0.02 significance level. (i.e. complete steps (a) to (e) similar to question 3) (e) Does your result from (d) contradict your result from (c)? Explain

Answer 1

a)

It is not mentioned that the women surveyed might not be selected randomly. This might lead to a bias in the observation as it will not represent the true population.

b)

Number of women who said that chocolate made them happier = 0.85*1708 = 1451.8 ~ 1452

c)

Given, p = 0.85

alpha = 0.02

Z_Critical = 2.33

Hence, 98% CI = p +/- Z_Critical * {p*(1-p)/n}^1/2 = 0.85 +/- 2.33 * {0.85*(1-0.85)/1708}^1/2

= {0.83,0.87}

Interpretation

There is 98% probability that the true proportion of women who said that chocolate made them happier lies in the interval {0.83,0.87}

d)

alpha = 0.02

Null and Alternate Hypothesis

H₀: µ = 0.8

H_a: µ > 0.8

Test Statistic

t = (p- µ₀)/ {p*(1-p)/n}^1/2 = (0.85- 0.8) / {0.85*(1-0.85)/1708}^1/2 = 5.79

p-value = TDIST(5.79,1708-1,1) = 0.000000004

Result

Since the p-value is less than 0.02, we reject the null hypothesis in favour of alternate hypothesis.

e)

Our result does not contradict result from part c