# AP423

1. Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative hypotheses, test statistic, p-value, critical value(s), and state the final conclusion that addresses the original claim.
– In a manual on how to have a number one song, it is stated that a song must be no longer to of 227.6 seconds and a standard deviation of 55.36 seconds. Use a 0.05 significance level and the accompanying minitab display to test the claim that the sample is from a population of songs with a mean greater than 210 seconds. What do these results suggest about the advice given in the manual?
Minitab display
One-sample T
Test of mu=210 vs >210
N=40
Mean = 227.60
St dev= 55.36
Se mean= 8.75
95% lower bound = 212.85
T= 2.01
P=0.026

2. A simple random sample of 59 adults is obtained from a normally distributed population, and each persons red blood cell count ( in cells per microliter) is measured. This sample mean is 5.25 and the sample standard deviation is 0.53. Use a 0.01 significance level and given calculator display to test the claim that the sample is from a population with a mean less than 5.4, which is a value often used for the upper limit of the range of normal values. What do the results suggest about the sample group?
t-test
u< 5.4
t= -2.173909174
p= 0.0169032278
x= 5.25
Sx= 0.53
N= 59
a. Identify the null and alternative hypotheses
b. Identify the statistic that is relevant to the test (type interger or decimal)
c. Identify the test statistic (round 3 decimal places)
d. Identify the p value (round 4 decimal places)
e. Find the critical values (left tailed,right or double)(round 3 decimal places)

3. For the following claim, find the null and alternative hypotheses, test statistic, p-value, critical value, and draw a conclusion. Assume that a simple random sample has been selected from a normally distributed population.
Claim: the mean IQ score of statistics professors is greater than 129.
Sample data: n=16
X=130
S=12
Significance level is a = 0.05