Question

(2 points) The sample mean and standard deviation from a random sample of 32 observations from a normal population were computed as a = 22 and s = 5. Calculate the t statistic of the test required to determine whether there is enough evidence to infer at the 7% significance level that the population mean is greater than 17. Test Statistic =

Answer 1

#1)

он
: µ = 17 vs
HO
: µ > 17

Given : = 22 , S = 5 , n = 32

T

Test statistic:

t =
X- s/n
= $\frac{22-17 }{5 /\sqrt{32}}$

= 5 / 0.8839

Test statistic t = 5.657

#2)

We can find the sample mean and standard deviation of the both data sets using excel function =AVERAGE( ) and =STDEV.S() respectively.

A B 11 8 2 2 15 8 33 40 6 50 6 6 2 8 70 9 8 4 9 92 9 10 3 8 11 12 =AVERAGE(A1:A10) !=AVERAGE(B1:B10) 13 1.7 7.6 14 -STDEV.S(A1:A10) =STDEV.S(B1:B10) 15 1.4181 1.4298 16

1
= 1.7 ,
S1
= 1.4181 , n₁ = 10 and  $//img.homeworklib.com/questions/e9eee840-f812-11ea-b80a-0714b8f1bb66.png?x-oss-process=image/resize,w_560$ = 7.6,
S2
= 1.4298, n₂ = 10

Confidence level = 0.921 ,Therefore α =1 - 0.921 = 0.079

degrees of freedom (df) = n₁+ n₂ - 2 = 10+10-2 = 18

We can find critical value using excel function =TINV( α , d.f )

=TINV( 0.079 , 18 ) = 1.862

Therefore critical value t = 1.862

Sp =$\sqrt{\frac{S_{1}^{2}(n_1-1)+S_{2}^{2}(n_2-1)}{n_1+n_2-2}}$

= $\sqrt{\frac{1.4181^{2}(10-1)+1.4298^{2}(10-1)}{10+10-2}}$

Sp = 1.4240

SE = $S_p*\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

= $1.4240*\sqrt{\frac{1}{10}+\frac{1}{10}}$

SE = 0.6368

Margin of error (E) = t*SE = 1.862*0.6368

E = 1.1858

(
1
- $1596138152561_blob.png$ ) = 1.7 - 7.6 = -5.9

Lower bound = (
1
- $1596138152561_blob.png$ ) - E = -5.9 - 1.1858

Lower bound = -7.0858

Upper bound = (
1
- $1596138152545_blob.png$ ) + E = -5.9 + 1.1858

Upper bound = -4.7142

Confidence interval = -7.0858 , -4.7142