# Math Help - Different Questions

1. ## Different Questions

Question 1:
• The mean and standard deviation for the quality grade-point averages of a random sample 36 college seniors are calculated to be 2.6 and 0.3, respectively. Find the 95% confidence interval for the mean of the entire senior class.

• In a random sample of 500 people eating lunch at a hospital cafeteria on various Fridays, it was found that x = 160 preferred seafood. Find a 90% confidence interval for the actual proportion of people who eat seafood on Fridays at this cafeteria.

Question 2:
• Hinton Press hypothesizes that the average life of its largest web press is 14,500 hours. They know that the standard deviation of press life is 2,100 hours. From a sample of 25 presses, the company finds a sample mean of 13,000 hours. At a 0.01 significance level, should the company conclude that the average life of the presses is less than the hypothesized 14,500 hours?

• An advertising executive thinks that the proportion of consumer who has seen his company’s advertising in news papers is 0.65 .The executive wants to estimate the consumer population proportion to with in ±0.05 and have 98 percent confidence in the estimate. How large a sample should be taken?

2. Originally Posted by aamirpk4
Question 1:
• The mean and standard deviation for the quality grade-point averages of a random sample 36 college seniors are calculated to be 2.6 and 0.3, respectively. Find the 95% confidence interval for the mean of the entire senior class.
Put:

$
t = \frac{\bar{X}-\mu}{(S/\sqrt{N})}
$

has a $t$ distribution with $\nu=N-1$ degrees of freedom.

Where $\bar{X}$ is the sample mean, $S$ is the sample SD, and $N$ is the sample size.

In this case $N=36$ so $\nu=35$, thus looking up the critical vaules of the $t$-distribution to give a $95\%$ interval for $t$ with $35$ degrees of freedom, we find that:

$t \in [-2.03, 2.03]$ with probability $95\%$.

So:

$[2.6-2.03 \times 0.3/\sqrt{35}, 2.6+2.03 \times 0.3/\sqrt{35} ]$

is the 95% confidence interval for the class mean.

RonL