Question

Listed in the data table are IQ scores for a random sample of subjects with medium lead levels in their blood. Also listed are statistics from a study done of IQ scores for a random sample of subjects with high lead levels. 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. Use a 0.01 significance level for both parts. Medium Lead Level 72 94 92 85 87 97 83 92 104 111 91 High Lead Level n2 = 11 x bar2 = 89.345 s2 = 10.173

The test statistic is

nothing.

​(Round to two decimal places as​ needed.)The​ P-value is

nothing.

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

State the conclusion for the test.

A.

Fail to rejectFail to reject

the null hypothesis. There

is notis not

sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores.

B.

RejectReject

the null hypothesis. There

isis

sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores.

C.

Fail to rejectFail to reject

the null hypothesis. There

isis

sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores.

D.

RejectReject

the null hypothesis. There

is notis not

sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores.

b. Construct a confidence interval suitable for testing the claim that the mean IQ scores for subjects with medium lead levels is higher than the mean for subjects with high lead levels.

nothingless than<mu 1μ1minus−mu 2μ2less than<nothing

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

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

▼

No,

Yes,

because the confidence interval contains

▼

zero.

only positive values.

only negative values.

Click to select your answer(s).

Answer 1

Solution:

Here, we have to use two sample t test for difference between two population means by assuming unequal population variances.

H₀: µ₁ = µ₂ versus H_a: µ₁ > µ₂

This is an upper tailed test.

Two sample t test for mean assuming unequal population variances

t = (X1bar – X2bar) / sqrt[(S₁² / n₁)+(S₂² / n₂)]

From given data, we have

X1bar = 91.636

X2bar = 89.345

S1 = 10.4330

S2 = 10.1730

n1 = 11

n2 = 11

df = 19

(by using excel and df formula)

(Degrees of freedom = [(S₁²/n₁) + (S₂²/n₂)]^2 / [((S₁²/n₁)^2/(n₁ – 1)) + ((S₂²/n₂)^2/(n₂ – 1))])

t = (91.636 - 89.345)/sqrt[(10.4330^2/11)+( 10.1730^2/11)]

t = 2.2910/4.3936

t = 0.5214

The test statistic is 0.52.

P-value = 0.3040

(by using t-table)

The​ P-value is 0.304.

P-value > α = 0.05

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that subjects with medium lead levels have higher IQ scores.

State the conclusion for the test.

Answer. A

Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores.

Part b

Confidence interval = (X1bar – X2bar) ± t*sqrt[(S₁² / n₁)+(S₂² / n₂)]

We assume confidence level = 95%

Critical t value = 2.0930

(X1bar – X2bar) = 2.2910

sqrt[(S₁² / n₁)+(S₂² / n₂)] = 4.3936

Confidence interval = (X1bar – X2bar) ± t*sqrt[(S₁² / n₁)+(S₂² / n₂)]

Confidence interval = 2.2910 ± 2.0930*4.3936

Confidence interval = 2.2910 ± 9.195805

Lower limit = 2.2910 - 9.195805 = -6.90481

Upper limit = 2.2910 + 9.195805 = 11.48681

Confidence interval = (-6.90481, 11.48681)

-6.90 < µ₁ - µ₂ < 11.49

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

Yes, because the confidence interval contains zero.