Respuesta :
Answer:
The data provide evidence that the mean number of lab classes taken by physics majors in January 1995 is not different from the mean number of lab classes taken in 2015.
Step-by-step explanation:
A t-test for difference between means can be used to determine whether the mean number of lab classes taken by physics majors in January 1995 is different from the mean number of lab classes taken in 2015.
The hypothesis is:
H₀: There is no difference between the two population means, i.e. μ₁ = μ₂.
Hₐ: There is a significant difference between the two population means, i.e. μ₁ ≠ μ₂.
The survey was conducted in a similar pattern. Assume that the two population standard deviations are same.
The information provided is:
[tex]\bar x_{1}=1.74\\s_{1}=1.49\\\bar x_{2}=1.87\\s_{1}=1.57\\n_{1}=n_{2}=150\\\alpha =0.10[/tex]
The test statistic is:
[tex]t=\frac{\bar x_{1}-\bar x_{2}}{s_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}[/tex]
Here s[tex]_{p}[/tex] is the pooled standard deviation.
Compute the value of pooled standard deviation as follows:
[tex]s_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}=\sqrt{\frac{(150-1)1.49^{2}+(150-1)1.57^{2}}{150+150-2}}=1.531[/tex]
Compute the test statistic value as follows:
[tex]t=\frac{\bar x_{1}-\bar x_{2}}{s_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}=\frac{1.74-1.87}{1.531\times\sqrt{\frac{1}{150}+\frac{1}{150}}}=-0.735[/tex]
The test statistic value is -0.735.
The decision rule is:
If the p - value of the test is less than the significance level, α = 0.10 then the null hypothesis will be rejected and vice-versa.
Compute the p-value as follows:
[tex]p-value=2P(t_{298}<-0.735)=0.463[/tex]
The p-value = 0.463 > α = 0.10.
As the p-value is more than the significance level the null hypothesis will not be rejected at 10% level of significance.
Conclusion:
The data provide evidence that the mean number of lab classes taken by physics majors in January 1995 is not different from the mean number of lab classes taken in 2015.
