Answer:
(a) It implies that if many samples of the members of the fitness club were selected and many 95% confidence intervals were constructed, the true proportion of all members of the fitness club who would quit would be captured by about 95% of the intervals.
(b) It implies that there is a 95% confidence that the true proportion of all members of the fitness club who would quit if there is an increase in the fee will be captures by the interval 0.105 – 0.225 (i.e. calculated as 0.18 – 0.075 = 0.105 and 0.18 + 0.075 = 0.225).
(c.) No, as there are plausible values above 20 as it is not all the intervals that are below 20. The implication of this is that there is a possibility that more than 20% would quit.
(d) The implication of the condition is that the distribution of the sample is approximately normal. This gives the opportunity to do estimations. However, the it will not be possible to estimate the 2 critical values if the distribution is not approximately normal. .
Explanation:
Note: This question is not complete. The complete question is therefore provided before answering the question as follows:
Members at a popular fitness club currently pay a $40 per month membership fee. The owner of the club wants to raise the fee to $50 but is concerned that some members will leave the gym if the fee increases. To investigate, the owner plans to survey a random sample of the club members and construct a 95% confidence interval for the proportion of all members who would quit if the fee was raised to $50.
(a) Explain the meaning of "95% confidence" in the context of this study.
(b) After the owner conducted the survey, he calculated the confidence interval to be 0.18 +/- 0.075. Interpret this interval in the context of the study.
(c.) According to the club’s accountant, the fee increase will be worthwhile if fewer than 20% of the members quit. According to the interval from part (b) can the owner be confident that the fee increase will be worthwhile? Explain.
(d) One of the conditions for calculating the confidence interval in part (b) is that np ≥ 10 and n(1 – p) ≥ 10. Explain why it is necessary to check this condition.
The explanation of the answer is now provided as follows:
(a) Explain the meaning of "95% confidence" in the context of this study.
It implies that if many samples of the members of the fitness club were selected and many 95% confidence intervals were constructed, the true proportion of all members of the fitness club who would quit would be captured by about 95% of the intervals.
(b) After the owner conducted the survey, he calculated the confidence interval to be 0.18 +/- 0.075. Interpret this interval in the context of the study.
It implies that there is a 95% confidence that the true proportion of all members of the fitness club who would quit if there is an increase in the fee will be captures by the interval 0.105 – 0.225 (i.e. calculated as 0.18 – 0.075 = 0.105 and 0.18 + 0.075 = 0.225).
(c.) According to the club’s accountant, the fee increase will be worthwhile if fewer than 20% of the members quit. According to the interval from part (b) can the owner be confident that the fee increase will be worthwhile? Explain.
No, as there are plausible values above 20 as it is not all the intervals that are below 20. The implication of this is that there is a possibility that more than 20% would quit.
(d) One of the conditions for calculating the confidence interval in part (b) is that np ≥ 10 and n(1 – p) ≥ 10. Explain why it is necessary to check this condition.
The implication of the condition is that the distribution of the sample is approximately normal. This gives the opportunity to do estimations. However, the it will not be possible to estimate the 2 critical values if the distribution is not approximately normal.