Answered

10.2 Hypothesis Testing for Population Mean

A recent survey reported that small businesses spend 21 hours a week marketing their business. A local chamber of commerce claims that small businesses in their area are not growing because these businesses are spending less than 21 hours a week on marketing. The chamber conducts a survey of 78 small businesses and finds that the average amount of time spent on marketing is 19.7 hours a week. Assuming that the population standard deviation is 5.1 hours, is there sufficient evidence to support the chamber of commerce's claim at the 0.05 level of significance?

Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blanks below.

[tex]\[
\begin{array}{l}
H_0: \mu = 21 \\
H_a: \mu \ \textless \ 21
\end{array}
\][/tex]



Answer :

GinnyAnswer
To solve this hypothesis testing problem, we can follow these detailed steps:

### Step 1: State the null and alternative hypotheses.
- Null Hypothesis (H₀): The null hypothesis is a statement that there is no effect or no difference, and it is given as:
- [tex]\( H_0: \mu = 21 \)[/tex]

- Alternative Hypothesis (Hₐ): The alternative hypothesis is the statement we want to test. It represents an effect, or a difference, and in this case, we are testing if the average time spent on marketing is less than 21 hours:
- [tex]\( H_a: \mu < 21 \)[/tex]

These hypotheses can be written as:
[tex]\[ H_0: \mu = 21 \][/tex]
[tex]\[ H_a: \mu < 21 \][/tex]

### Step 2: Compute the test statistic.
- Sample Size (n): 78
- Sample Mean ( [tex]\(\bar{x} \)[/tex]): 19.7
- Population Mean ( [tex]\(\mu\)[/tex] ): 21
- Population Standard Deviation (σ): 5.1
- Significance Level (α): 0.05

The test statistic for this hypothesis test is the z-score, calculated using the formula:
[tex]\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \][/tex]

Plugging in the numbers:
[tex]\[ z = \frac{19.7 - 21}{\frac{5.1}{\sqrt{78}}} \approx -2.251 \][/tex]

### Step 3: Calculate the p-value.
The p-value is the probability of obtaining a test statistic at least as extreme as the one that was observed, assuming that the null hypothesis is true. For this left-tailed test:
[tex]\[ \text{p-value} \approx 0.0122 \][/tex]

### Step 4: Compare p-value with the significance level.
- Significance Level (α): 0.05
- P-value: 0.0122

If the p-value is less than the significance level (0.05), we reject the null hypothesis. Here, 0.0122 < 0.05, which means we reject the null hypothesis.

### Conclusion:
Since the p-value is less than the significance level, there is sufficient evidence to support the claim of the chamber of commerce that small businesses in their area spend less than 21 hours a week on marketing.

In summary:
- Null Hypothesis: [tex]\( H_0: \mu = 21 \)[/tex]
- Alternative Hypothesis: [tex]\( H_a: \mu < 21 \)[/tex]
- Test Statistic (z-score): -2.251
- P-value: 0.0122
- Decision: Reject [tex]\( H_0 \)[/tex]

There is sufficient evidence to support the chamber of commerce's claim that small businesses in their area are spending less than 21 hours a week on marketing.