Q:

Fifty specimens of a new computer chip were tested for speed in a certain application, along with 50 specimens of chips with the old design. The average speed, in MHz, for the new chips was 495.6, and the standard deviation was 19.4. The average speed for the old chips was 481.2, and the standard deviation was 14.3.Can you conclude that the mean speed of the new chips is greater than that of the old chips?

Accepted Solution

A:
Answer: [tex] t= \frac{495.6-481.2}{\sqrt{\frac{19.4^2}{50} +\frac{14.3^2}{50}}}[/tex][tex]p_v =2*P(t_{(98)}>4.224)=0.0000538[/tex]So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the two meansStep-by-step explanation:Data given and notation[tex]\bar X_{new}= 495.6[/tex] represent the mean for the sample new[tex]\bar X_{old}=481.2[/tex] represent the mean for the sample old[tex]s_{new}=19.4[/tex] represent the sample standard deviation for the sample new[tex]s_{old}=14.3[/tex] represent the sample standard deviation for the sample old[tex]n_{new}=50[/tex] sample size for the group new[tex]n_{old}=50[/tex] sample size for the group oldt would represent the statistic (variable of interest)Concepts and formulas to useWe need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:H0:[tex]\mu_{new}=\mu_{old}[/tex]H1:[tex]\mu_{new} \neq \mu_{old}[/tex]If we analyze the size for the samples both are > 30 so for this case but we don't know the population deviations, so is better apply a t test to compare meanst-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.Calculate the statisticAnd with this we can replace in formula (1) like this: [tex] t= \frac{495.6-481.2}{\sqrt{\frac{19.4^2}{50} +\frac{14.3^2}{50}}}[/tex]  Statistical decisionFor this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:[tex]df=n_{new}+n_{old}-2=50+50-2=98[/tex]Since is a bilateral test the p value would be:[tex]p_v =2*P(t_{(98)}>4.224)=0.0000538[/tex]So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the two means