Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1. Test appropriate hypothesis.

#### Solution

Given that $n_1 = 150$, $X_1= 15$, $n_2=150$ and $X_2=9$.

The sample proportions are

$\hat{p}_1=\frac{X_1}{n_1}=\frac{15}{150}=0.1$.

$\hat{p}_2=\frac{X_2}{n_2}=\frac{9}{150}=0.06$.

The pooled estimate of sample proportion is

$\hat{p} =\frac{X_1+X_2}{n_1+n_2}=\frac{15+9}{150+150} =0.08$

#### Step 1 State the hypothesis testing problem

The hypothesis testing problem is

$H_0 : p_1 = p_2$ against $H_1 : p_1 > p_2$ ($\textit{right-tailed}$)

#### Step 2 Define test statistic

The test statistic for testing above hypothesis testing problem is

` $$ \begin{aligned} Z & =\frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}. \end{aligned} $$ `

The test statistic $Z$ follows standard normal distribution $N(0,1)$.

#### Step 3 Specify the level of significance $\alpha$

The significance level is $\alpha = 0.05$.

#### Step 4 Determine the critical value

As the alternative hypothesis is $\textit{right-tailed}$, the critical value of $Z$ $\text{ is }$ $\text{1.64}$ (From Normal Statistical Table).

The rejection region (i.e. critical region) is $\text{Z > 1.64}$.

#### Step 5 Computation

The test statistic under the null hypothesis is

` $$ \begin{aligned} Z_{obs}&= \frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}\\ &= \frac{(0.1-0.06)-0}{\sqrt{\frac{0.08*(1-0.08)}{150}+\frac{0.08*(1-0.08)}{150}}}\\ &= 1.277 \end{aligned} $$ `

#### Step 6 Decision

Traditional approach:

The rejection region (i.e. critical region) is $\text{Z > 1.64}$. The test statistic is $Z_{obs} =1.277$ which falls $outside$ the critical region, we $\textit{fail to reject}$ the null hypothesis.

OR

$p$-value approach:

The test is $\text{right-tailed}$ test, so the p-value is the area to the $\text{right}$ of the test statistic ($Z_{obs}=1.277$) is p-value = $0.1008$.

The p-value is $0.1008$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.

We conclude that there is no sufficient evidence to reject the null hypothesis, so the data do not show that OS2 has fewer system failures than OS1.