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Answer:
It does not appear that, as a group, the students are reasonably good at estimating one minute.
Step-by-step explanation:
We are given the following data in the question:
75, 88, 51, 73, 49, 31, 69, 74, 72, 59, 72, 81, 99, 101, 73
Formula:
where are data points,
is the mean and n is the number of observations.
Sum of squares of differences = 4739.733
Population mean, μ = 60 minutes
Sample mean, = 71.13 minutes
Sample size, n = 15
Alpha, α = 0.10
Sample standard deviation, s = 18.39 minutes
First, we design the null and the alternate hypothesis
We use Two-tailed t test to perform this hypothesis.
Formula:
Putting all the values, we have
Calculating the p-value from the table, we have,
P-value = 0.034354
Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Thus, we conclude that it does not appear that, as a group, the students are reasonably good at estimating one minute.