The number of visitors (\(N\)) to a museum per day is recorded over a period of 180 days. The results are summarized in the following table:
| Number of Visitors (\(N\)) |
Frequency |
| 100 \(\le N <\) 150 |
15 |
| 150 \(\le N <\) 200 |
35 |
| 200 \(\le N <\) 250 |
58 |
| 250 \(\le N <\) 300 |
42 |
| 300 \(\le N <\) 350 |
21 |
| 350 \(\le N <\) 400 |
9 |
(a) Estimate the mean number of visitors per day. [2]
(b) Estimate the standard deviation of the number of visitors per day. [1]
(c) Find an estimate for the median number of visitors per day. [2]
(d) The museum curator claims that the mean number of visitors is 240. Test this claim at the 5% significance level, stating your hypotheses clearly. Assume the number of visitors follows a normal distribution with the standard deviation found in part (b). [4]
Question
Estimate the mean number of visitors per day. [2]
Hint
Hint
Hint
Use the mid-interval values (e.g. 125 for 100 ≤ N < 150) and multiply by the corresponding frequencies. Sum the products and divide by 180.
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Approximately 234 visitors
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Question
Estimate the standard deviation of the number of visitors per day. [1]
Hint
Hint
Hint
Use the formula for the sample standard deviation with the midpoints as data values.
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Approximately 63.9 visitors
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Find an estimate for the median number of visitors per day. [2]
Hint
Hint
Hint
Identify the median class by using cumulative frequencies; the median is the value at the 90th data point.
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Approximately 231 visitors
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The museum curator claims that the mean number of visitors is 240. Test this claim at the 5% significance level, stating your hypotheses clearly. [4]
Hint
Hint
Hint
State H₀: \(\mu = 240\) and H₁: \(\mu \neq 240\). Calculate the test statistic using \(t = \frac{(\bar{x} – 240)}{(s/\sqrt{180})}\) with \(\bar{x} \approx 234\) and \(s \approx 63.9\), then compare the p-value to 0.05.
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H₀: \(\mu = 240\); H₁: \(\mu
eq 240\). The computed t-statistic is approximately -1.17 with a p-value of about 0.243, so we do not reject H₀.
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