Question

(a) According to a certain chemical theory, the percentage of iron in a certain compound should be 12.1. To test this theory, it was decided to analyze 9 different random samples of the compound to see if the measurements would differ significantly from 12.1%. Suppose the 9 samples gave the following results:

11.7 12.2 10.9 11.4 11.3 12.0 11.1 10.7 11.6

Use either classical or p-value approach to verify the hypothesis at level of significance a= 0.05. Assume that the population is normally distributed.

(b) A bank manager estimates that 75% of the customers wait less than ten minutes in the queue at the bank. To look into this estimation, a customer group conducted a survey with a sample of 90 customers. It was found that 15% of the customers waited at least ten minutes in the queue. At the level of significance a= 0.01, is there enough evidence to conclude that more than 75% of the customers wait less than ten minutes in the queue at the bank?

Answer 1

a.

For this question, we shall use the classical approach.

We first determine the value of the sample mean and standard deviation.

The mean is given by,

$\bar{x}=\sum(x)/n$ where $n=9$

$\bar{x}=102.9/9=11.43333333$

To find the standard deviation, we need to find the variance given by the formula,

$V(X)=(\sum(x^2)-(\sum(x))^2/n)/(n-1)$

$V(X)=(1178.45-1176.49)/8=0.245$

The standard deviation $SD(X)=\sqrt{V(X)}=\sqrt{0.245}=0.494974747$

The hypotheses tested are,

$H_0:\mu=12.1\space Against\space H_1:\mu\not=12.1$

To perform this test, we shall apply the student's t distribution since the population variance is not known and the sample size is less than 30.

The test statistic is given as,

$t^*=(\bar{x}-\mu)/(SD(X)/\sqrt{n})$

$t^*=(12.1-11.4333)/(0.494974747/\sqrt{9})=0.666667/0.164992=4.04061$

$t^*$ is compared with the table value at $\alpha=0.05$ with $(n-1)=9-1=8$ degrees of freedom.

Thus, $t_{\alpha/2,8}=t_{0.05/2,8}=t_{0.025,8}=2.306$, and the null hypothesis is rejected if $t^*\gt t_{0.025,8}$

Since $t^*=4.04061\gt t_{0.025,8}=2.306$, we reject the null hypothesis and we conclude that there is sufficient evidence to show that the measurements differ significantly from 12.1% at 5% significance level.

b.

For this question, we test hypothesis on a single population proportion.

We test the following hypotheses,

$H_0:p=0.75$

$Against$

$H_1:p\gt 0.75$

From the information above, the proportion estimate $(\hat{p})$ is 85%=0.85. We shall apply the z-test to perform this hypothesis test since $\hat{p}\sim N(p,(p*(1-p))/n)$ where $p=0.75,n=90$

Now, the test statistic is given as,

$Z=(\hat{p}-p)/\sqrt{p*(1-p)/n}=(0.85-0.75)/\sqrt{(0.75*0.25)/90}=0.1/0.04564=2.1989(4dp)$

$Z$ is compared with the table value at $\alpha=0.01$.

The table value is,

$Z_{\alpha}=Z_{0.01}=2.33$ and the null hypothesis is rejected if $Z\gt Z_\alpha$

Since $Z=2.1989\lt Z_\alpha=2.33$, we fail to reject the null hypothesis and conclude that there is no enough evidence to show that more than 75% of the customers wait less than ten minutes in the queue at the bank at 1% level of significance.