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 0.05 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?

72

83

41

64

39

22

60

64

67

51

67

71

92

91

62

Assuming all conditions for conducting a hypothesis test are​ met, what are the null and alternative​ hypotheses?

A. Upper H0​: μ=60 seconds

Upper H1​: μ<60 seconds

B. Upper H0​: μ=60 seconds

Upper H1​: μ>60 seconds

C.Upper H0​: μ=60 seconds

Upper H1​:μ≠60 seconds

D. Upper H0​: μ≠60 seconds

Upper H1​: μ=60 seconds

.

Determine the test statistic.

___________ ​(Round to two decimal places as​ needed.)

Determine the​ P-value.

_________________ ​(Round to three decimal places as​ needed.)

State the final conclusion that addresses the original claim.

▼ FAILED TO REJECT / REJECT Upper H0. 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 ▼ does not appear / appears ​that, as a​ group, the students are reasonably good at estimating one minute.

Answer 1

# Q ) 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 0.05 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?

Answer :

We have given :   

μ = population mean = 60 seconds

α = level of significance = 0.05

n = sample size = 15

Claim :

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?

Assuming all conditions for conducting a hypothesis test are​ met, what are the null and alternative​ hypotheses?

C ) Upper H0 ​: μ = 60 seconds

Upper H1​ : μ ≠ 60 seconds

## Determine the test statistic. t = ( x̄ - μ ) * √ n / s

where x̄ = sample mean and s = sample standard deviation

here first we need to calculate sample mean and sample standard deviation .

x̄ = Σ x̄ / n

= ( 72 + 83 + 41 + 64 + 39 + 22 + 60 + 64 + 67 + 51 + 67 + 71 + 92 + 91 + 62 ) / 15 = 946 / 15

= 63.0666

s = √ ( Σ ( x -   x̄ ) ² / ( n -1 ))

  Σ ( x -   x̄ ) ² / ( n -1) = sample variance

( (72 - 63.0666 ) ² + ( 83 - 63.0666)² + ( 41 - 63.0666) ²  + ( 64 - 63.0666) ² + ( 39 - 63.0666) ² +

( 22 - 63.0666 ) ²  + ( 60 - 63.0666) ² + ( 64 - 63.0666) ² + ( 67 - 63.0666) ² + ( 51 - 63.0666) ² +

( 67 - 63.0666) ² + ( 71 - 63.0666) ² + ( 92 - 63.0666) ² + ( 91 - 63.0666) ² + ( 62 - 63.0666)²  ) / 14

= ( 79.8044 + 397.3378 + 486.9378 + 0.87111 + 579.2044 + 1686.471 + 9.4044 + 0.87111 + 15.4711 +

145.6044 + 15.4711 + 62.9377 + 837.1378 + 780.2711 + 1.137778 ) / 14

= 5098.933 / 14

= 364.2095

Therefore , s = √ 364.2095

= 19.0843

t = ( 63.0666 - 60 ) * √ 15 / 19.0843

t = 11.87715 /  19.0843  

= 0.622353 ie

= 0.62 ​(Round to two decimal places as​ needed.)

## Determine the​ P-value. :

it is two tailed test : and df = degree of freedom = n -1 = 14

p value = 2 * P [ t > 0.62 ] now using statistical table

= 2 * 0.272609 = 0.545218

= 0.545  ​(Round to three decimal places as​ needed.)

Decision :

we reject Ho if p value is less than α value using p value approach here p value is greater than  

α value here we fail to reject Ho at given level of significance .

State the final conclusion that addresses the original claim.

FAILED TO REJECT Upper H0. There is not sufficient evidence to conclude that the original claim that the mean of the population of estimates is 60 seconds is not correct. It does not appear that, as a​ group, the students are reasonably good at estimating one minute.