Question

9. Listed in the data table are amounts of​ strontium-90 (in​ millibecquerels, or​ mBq, per gram of​ calcium) in a simple random sample of baby teeth obtained from residents in two cities. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Complete parts​ (a) and​ (b) below.

City_#1 City_#2

104         117

86           92

121         100

114         85

101         84

104         107

213         110

144         111

290         135

100         133

306         101

145         209

The test statistic is ____

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

The​ P-value is ____

(Round to three decimal places as​ needed.)

State the conclusion for the test.

A.Fail to reject the null hypothesis. There is sufficient evidence to support the claim that archbishops have lower mean longevity than monarchs.

B.Reject the null hypothesis. There is not sufficient evidence to support the claim that archbishops have lower mean longevity than monarchs.

C.Reject the null hypothesis. There is sufficient evidence to support the claim that archbishops have lower mean longevity than monarchs.

D.Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that archbishops have lower mean longevity than monarchs.

b. Construct a confidence interval suitable for testing the claim that the mean longevity for archbishops is less than the mean for monarchs after coronation.

___<µ1-µ2<____

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

Does the confidence interval support the conclusion of the​ test?

(No,Yes,) because the confidence interval contains (only negative values., zero, only positive values.)

Answer 1

City 1	City 2
104	117
86	92
121	100
114	85
101	84
104	107
213	110
144	111
290	135
100	133
306	101
145	209

In this question, we will take the 2-sample T-Test using unequal variances. I'm assuming that the motive behind this test is to see if there is any statistically significant difference between the amounts of strontium-90 in the baby teeth of the kids of both the cities. Therefore, we will take the hypothesized mean difference to be equal to 0 i.e.

The null hypothesis will be given as the average amount of strontium-90 in the baby teeth of the children of both the cities is equal i.e. ₁-₂ = 0.

Alternate hypothesis: ​​​​​​​₁-₂ 0. Therefore, it's a two-tail T-test.

Though their are formulas to solve this T-test for unequal variances, we can simply use Excel to solve it for us. In Excel, go to Data Solver > T-Test assuming unequal variances > Enter the data variables, keep the hypothesized mean difference as 0 and select the value of ​​​​​​ = 0.05.

You will get the following result:

t-Test: Two-Sample Assuming Unequal Variances
	City 1	City 2
Mean	152.3333333	115.3333333
Variance	5751.515152	1134.424242
Observations	12	12
Hypothesized Mean Difference	0
df	15
t Stat	1.544581216
P(T<=t) one-tail	0.071640256
t Critical one-tail	1.753050356
P(T<=t) two-tail	0.143280511
t Critical two-tail	2.131449546

Here, you can see that the T-statistic for the data is 1.545(rounded) and the p-value is 0.14(since this is a two-tail test).

Thus, the result of this test is that there is no statistically significant evidence that the null hypothesis is true i.e. we fail to reject the null hypothesis. In other words, you can say that there is no statistically significant evidence that the average amount of strontium-90 in the baby teeth of the populations in the two cities is equal.