Question

A group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are listed below. Use a 0.10 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one​ minute? 75 88 51 73 49 31 69 74 72 59 72 81 99 101 73 What are the null and alternative​ hypotheses? A. Upper H 0​: muequals60 seconds Upper H 1​: munot equals60 seconds B. Upper H 0​: munot equals60 seconds Upper H 1​: muequals60 seconds C. Upper H 0​: muequals60 seconds Upper H 1​: muless than60 seconds D. Upper H 0​: muequals60 seconds Upper H 1​: mugreater than60 seconds Determine the test statistic. nothing ​(Round to two decimal places as​ needed.) Determine the​ P-value. nothing ​(Round to three decimal places as​ needed.) State the final conclusion that addresses the original claim. ▼ Fail to reject Reject Upper H 0. There is ▼ sufficient not sufficient evidence to conclude that the original claim that the mean of the population of estimates is 60 seconds ▼ is is not correct. It ▼ appears does not appear ​that, as a​ group, the students are reasonably good at estimating one minute.

Answer 1

It does not appear ​that, as a​ group, the students are reasonably good at estimating one minute.

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:

`\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}`

where
`x_i` are data points,
`\bar{x}` is the mean and n is the number of observations.

`Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}`

`Mean =\displaystyle(1067)/(15) = 71.13`

Sum of squares of differences = 4739.733

`S.D = \sqrt{(4739.733)/(14)} = 18.39`

Population mean, μ = 60 minutes

Sample mean,
`\bar{x}` = 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

`H_(0): \mu = 60\text{ minutes}\\H_A: \mu \\eq 60\text{ minutes}`

We use Two-tailed t test to perform this hypothesis.

Formula:

`t_(stat) = \displaystyle\frac{\bar{x} - \mu}{(s)/(√(n)) }`

Putting all the values, we have

`t_(stat) = \displaystyle(71.13 - 60)/((18.39)/(√(15)) ) = 2.34`

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.