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Math  /  Data & Statistics

QuestionGuestion 8 of 80 This topti 20 pointo paspole Suborif the σ12=σ22\sigma_{1}^{2}=\sigma_{2}^{2}. Use a 0 o 0s n1=14,n2=12,x1=17,x2=18,x1=2.5,32=2.8n_{1}=14, n_{2}=12, x_{1}=17, x_{2}=18, x_{1}=2.5,3_{2}=2.8

Studdy Solution

STEP 1

1. The samples are randomly selected from normal populations.
2. The population variances are equal: σ12=σ22 \sigma_1^2 = \sigma_2^2 .
3. The significance level is α=0.05 \alpha = 0.05 .
4. We are testing the null hypothesis H0:μ1=μ2 H_0: \mu_1 = \mu_2 against the alternative hypothesis Ha:μ1≠μ2 H_a: \mu_1 \neq \mu_2 .

STEP 2

1. Calculate the pooled standard deviation.
2. Calculate the test statistic.
3. Determine the critical values for the t-distribution.
4. Compare the test statistic to the critical values.

STEP 3

Calculate the pooled standard deviation sp s_p using the formula:
sp=(n1−1)s12+(n2−1)s22n1+n2−2s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}
Substitute the given values:
sp=(14−1)(2.5)2+(12−1)(2.8)214+12−2s_p = \sqrt{\frac{(14 - 1)(2.5)^2 + (12 - 1)(2.8)^2}{14 + 12 - 2}}
Calculate:
sp=13×6.25+11×7.8424s_p = \sqrt{\frac{13 \times 6.25 + 11 \times 7.84}{24}}
sp=81.25+86.2424s_p = \sqrt{\frac{81.25 + 86.24}{24}}
sp=167.4924s_p = \sqrt{\frac{167.49}{24}}
sp≈6.97875≈2.641s_p \approx \sqrt{6.97875} \approx 2.641

STEP 4

Calculate the test statistic t t using the formula:
t=xˉ1−xˉ2sp1n1+1n2t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}
Substitute the given values:
t=17−182.641114+112t = \frac{17 - 18}{2.641 \sqrt{\frac{1}{14} + \frac{1}{12}}}
Calculate:
t=−12.6410.0714+0.0833t = \frac{-1}{2.641 \sqrt{0.0714 + 0.0833}}
t=−12.6410.1547t = \frac{-1}{2.641 \sqrt{0.1547}}
t=−12.641×0.3933t = \frac{-1}{2.641 \times 0.3933}
t≈−11.038≈−0.963t \approx \frac{-1}{1.038} \approx -0.963

STEP 5

Determine the critical values for a two-tailed test with α=0.05 \alpha = 0.05 and degrees of freedom df=n1+n2−2=24 df = n_1 + n_2 - 2 = 24 .
Using a t-distribution table or calculator, find the critical values t0.025,24 t_{0.025, 24} .
t0.025,24≈±2.064t_{0.025, 24} \approx \pm 2.064

STEP 6

Compare the test statistic to the critical values:
Since −2.064<−0.963<2.064 -2.064 < -0.963 < 2.064 , we fail to reject the null hypothesis H0:μ1=μ2 H_0: \mu_1 = \mu_2 .
The conclusion is that there is not enough evidence to reject the claim that the population means are equal at the 0.05 significance level.

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