Math  /  Data & Statistics

QuestionSydnel Thomas 11/22/244.00PM11 / 22 / 244.00 \mathrm{PM} Question 5 of 20 mileste20 pint(0) possiblo Thil question Ipolnt(s) possib- \mathrm{I}_{\text {polnt(s) possib- }} Submil test
Find the standardized lest statistic, LL, to test the claim that μ1>μ2\mu_{1}>\mu_{2}. Two samples are randomly selected and come from populations that are normat. The samplo statistics are given below Aasume that σ12σ22\sigma_{1}^{2} \neq \sigma_{2}^{2}. n1=18,n2=13,xˉ1=485,xˉ2=470,s1=40,s2=25n_{1}=18, n_{2}=13, \bar{x}_{1}=485, \bar{x}_{2}=470, s_{1}=40, s_{2}=25

Studdy Solution

STEP 1

1. The samples are independent and come from normally distributed populations.
2. The population variances are not equal (σ12σ22\sigma_1^2 \neq \sigma_2^2).
3. We are using a two-sample t-test to compare the means.

STEP 2

1. Calculate the standard error of the difference between the sample means.
2. Compute the test statistic LL.

STEP 3

Calculate the standard error of the difference between the sample means using the formula:
SE=s12n1+s22n2SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}
Substitute the given values:
SE=40218+25213SE = \sqrt{\frac{40^2}{18} + \frac{25^2}{13}}
Calculate each term:
40218=16001888.89\frac{40^2}{18} = \frac{1600}{18} \approx 88.89
25213=6251348.08\frac{25^2}{13} = \frac{625}{13} \approx 48.08
Add these values:
SE=88.89+48.08=136.9711.70SE = \sqrt{88.89 + 48.08} = \sqrt{136.97} \approx 11.70

STEP 4

Compute the test statistic LL using the formula:
L=xˉ1xˉ2SEL = \frac{\bar{x}_1 - \bar{x}_2}{SE}
Substitute the given values:
L=48547011.70=1511.701.28L = \frac{485 - 470}{11.70} = \frac{15}{11.70} \approx 1.28
The standardized test statistic LL is:
1.28 \boxed{1.28}

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