Question

1. Two researchers, Beth and Frank, ask a random sample of teenagers whether or not they had been to the movies in the last month. They find that their sample proportion (p) of teens who said yes was .8 (80%) When they construct a confidence interval based on this sample proportion, Beth comes up with (.73, .87) while Frank gets (.75, .89). Indicate which interval has to be wrong, and explain your choice.

2. In a random sample of 28 families, the average weekly food expense was $95.60 with a standard deviation of $22.50. Determine whether a normal distribution or a t-distribution should be used or whether neither of these can be used to construct a confidence interval. Explain your answer. Assume the distribution of weekly food expenses is normally shaped.

Answer 1

(1) the upper and lower limit of confidence interval is equidistant from the sample value.

i,e. confidence interval=sample value margin of error

lower limit=sample value - margin of error

upper limit=sample value+margin of error

so the interval of Beth comes up with (.73, .87) is equidistrant form the sample value=0.8,

that is the sample value=(lower limit +upper limit)/2=(0.87+0.73)/2=0.8

with margin of error=(upper limit-lower limit)/2=(0.87-0.73)/2=0.07

interval of Frank (0.75,0.89) is wrong as its sample value=(0.89+0.75)/2=0.82 is not equal to sample value=0.8

(2) since sample size =28 is less than 30 and population standard deviation is not given, so we use t-distribution .