In hypothesis testing, we test about a population parameter. In this example, we are testing population mean. Let's see a real life problem followed by its solution.

A marketing company has suggested that the cost of monthly cable TV subscription has risen dramatically, which is causing more people to use illegal satellite dishes. Cable TV companies claim that their full cable package subscriptions cost on average $\$$50 a month. The marketing company wants to demonstrate that the cost is significantly greater than $\$$50 and randomly selects 40 cable TV subscribers and determines the price they pay for their monthly cable. When the sample of n = 40 subscriptions is randomly selected, the mean of the cable costs is calculated to be $\bar x$ = 50.575. Assuming that $\sigma$ is known to equal 1.65, test whether the marketing company's claim is true.

Null hypothesis, H$_o$: $\mu$ = 50

Alternative hypothesis, H$_1$: $\mu$ > 50

We set $\alpha$ = 5% or 0.05 i.e. we do not wish to mistake more than 5%.

TS or Z = $\frac{\bar x - \mu}{\sigma/\sqrt n}$

As the sample size is n = 40 ($\gt$ 30), we may assume, according to Central Limit Theorem (CLT) that TS ~ N(0,1), when null hypothesis is true.

From the problem, we have,

$\bar x = \$50.575$

Population mean, $\mu = \$50$

Population standard deviation, $\sigma$ = 1.65

and sample size, n =40.

Hence,

$\begin{align}TS, Z = \frac{(50.575-50)}{1.65/\sqrt 40}\\

& = 2.204

\end{align}$

Since H$_1: \mu>50$

$\begin{align}p-value = P[Z>2.204] \\

& = 1- 0.9875\\

& = 0.0125

\end{align}$

As p-value $\lt \alpha \;(=0.05)$, we may reject the null hypothesis at 0.05 level of significance.

We have enough evidence to claim that the cost of cable TV subscription has risen.

The above method is called p-value approach. However, we can employ another method called critical region approach.

The first three steps are same.

We assumed $\alpha$ = 0.05. Since $H_1 \gt$ 50, the rejection region is Z$\ge Z_\alpha$.

For $\alpha = 0.05, Z_\alpha = 1.645. $

Therefore, reject $H_o if Z \ge1.645$.

We have already calculated that, Z = 2.204

Since Z$\ge Z_\alpha$, we may reject the null hypothesis ($H_o$) at 5% level of significance.

We have enough evidence to claim that the cost of cable TV subscription has risen.

**Problem**:A marketing company has suggested that the cost of monthly cable TV subscription has risen dramatically, which is causing more people to use illegal satellite dishes. Cable TV companies claim that their full cable package subscriptions cost on average $\$$50 a month. The marketing company wants to demonstrate that the cost is significantly greater than $\$$50 and randomly selects 40 cable TV subscribers and determines the price they pay for their monthly cable. When the sample of n = 40 subscriptions is randomly selected, the mean of the cable costs is calculated to be $\bar x$ = 50.575. Assuming that $\sigma$ is known to equal 1.65, test whether the marketing company's claim is true.

**Solution**:**Step 1:**We set up the hypotheses:Null hypothesis, H$_o$: $\mu$ = 50

Alternative hypothesis, H$_1$: $\mu$ > 50

**Step 2:**Desired level of significanceWe set $\alpha$ = 5% or 0.05 i.e. we do not wish to mistake more than 5%.

**Step 3:**Test statistic (TS):TS or Z = $\frac{\bar x - \mu}{\sigma/\sqrt n}$

As the sample size is n = 40 ($\gt$ 30), we may assume, according to Central Limit Theorem (CLT) that TS ~ N(0,1), when null hypothesis is true.

**Step 4:**Value of TSFrom the problem, we have,

$\bar x = \$50.575$

Population mean, $\mu = \$50$

Population standard deviation, $\sigma$ = 1.65

and sample size, n =40.

Hence,

$\begin{align}TS, Z = \frac{(50.575-50)}{1.65/\sqrt 40}\\

& = 2.204

\end{align}$

**Step 5:**p-valueSince H$_1: \mu>50$

$\begin{align}p-value = P[Z>2.204] \\

& = 1- 0.9875\\

& = 0.0125

\end{align}$

**Step 6:**DecisionAs p-value $\lt \alpha \;(=0.05)$, we may reject the null hypothesis at 0.05 level of significance.

**Step 7**: InterpretationWe have enough evidence to claim that the cost of cable TV subscription has risen.

**Alternative solution:**The above method is called p-value approach. However, we can employ another method called critical region approach.

The first three steps are same.

**Step 4:**Construct the rejection region (also called critical region)We assumed $\alpha$ = 0.05. Since $H_1 \gt$ 50, the rejection region is Z$\ge Z_\alpha$.

For $\alpha = 0.05, Z_\alpha = 1.645. $

Therefore, reject $H_o if Z \ge1.645$.

**Step 5**: Value of ZWe have already calculated that, Z = 2.204

**Step 6:**DecisionSince Z$\ge Z_\alpha$, we may reject the null hypothesis ($H_o$) at 5% level of significance.

**Step 7:**InterpretationWe have enough evidence to claim that the cost of cable TV subscription has risen.

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