Question

1. According to an educational report, the amount of time that students spend “off-task” (such as by checking their phones) during a one-hour lecture is approximately normally distributed with a mean of 3.2 minutes and a standard deviation of 2.7 minutes. An educator is interested in determining at the α = 0.05 level if the average amount of time his students spend on their phones while he is lecturing differs from the value given in the journal. During a particular class period in which he has 37 students, he noted that the average amount of time students spent on their phones was 4.2 minutes.

(a) State the null and alternative hypotheses for this test.

(b) Compute an appropriate test statistic.

(c) Determine the p-value for this test.

(d) State, in words, your conclusion.

2. Construct a 95% confidence interval for the mean amount of time students spend on their phones. Does this confidence interval support your conclusion from the hypothesis test in part (d)? Why or why not?

Answer 1

a)

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 3.2
Alternative Hypothesis: μ ≠ 3.2

b)

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (4.2 - 3.2)/(2.7/sqrt(37))
z = 2.25

c)

P-value Approach
P-value = 0.0244

d)

As P-value < 0.05, reject the null hypothesis.
There is sufficient evidence to conclude that average amount of time his students spend on their phones while he is lecturing differs from the value given in the journal.

2)

sample mean, xbar = 4.2
sample standard deviation, σ = 2.7
sample size, n = 37

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 2.7/sqrt(37)
ME = 0.87

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (4.2 - 1.96 * 2.7/sqrt(37) , 4.2 + 1.96 * 2.7/sqrt(37))
CI = (3.33 , 5.07)

yes this confidence interval support your conclusion from the hypothesis test in part (d)
because confidence interval does not contain 3.2 minutes