Question

Score: 0.43 of 1 pt 13 of 16 (16 cm w ocure. 00.170, 1. TUTUP % 8.3.22-T Question Help 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? 68 8440664221 63 62 67 48 62 7091 89 65 What are the null and alternative hypotheses? O A. Ho 60 seconds H <60 seconds OC. Ho 60 seconds Hyu 60 seconds OB. Hou = 60 seconds H > 60 seconds OD. Ho 60 seconds H 60 seconds

Answer 1

Solution:

The null and alternative hypotheses are given as below:

Answer: D.

H0: µ = 60 seconds

H1: µ ≠ 60 seconds

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 60

Xbar = 62.53333333

S = 18.82576755

n = 15

df = n – 1 = 14

α = 0.10

Critical value = - 1.7613 and 1.7613

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (62.53333333 – 60)/[ 18.82576755/sqrt(15)]

t = 0.5212

P-value = 0.6104

(by using t-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the students are reasonably good at estimating one minute.