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On average, teenagers spend 4.5 hours per week on the phone. The organization thinks this average is currently true. The population standard deviation is 1.4 hours. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0 hours. Conduct a hypothesis test at the 0.05 level of significance.

Question 1 (1 point)

What is the null hypothesis?

[tex]\[ H_0: \mu = 4.5 \][/tex]

Sagot :

GinnyAnswer
Sure, let's go through the hypothesis testing step-by-step.

### Step 1: Setting up the Hypotheses

Null Hypothesis (H₀): The null hypothesis is that the population mean (μ) is equal to 4.5 hours. This can be written as:
[tex]\[ H₀: \mu = 4.5 \][/tex]

Alternative Hypothesis (H₁): The alternative hypothesis is that the population mean (μ) is not equal to 4.5 hours. This can be written as:
[tex]\[ H₁: \mu \neq 4.5 \][/tex]

### Step 2: Gathering Data and Known Values
- Sample size (n): 15
- Sample mean ( [tex]\(\bar{x} \)[/tex] ): 4.75
- Population mean (μ): 4.5
- Sample standard deviation (s): 2.0
- Population standard deviation (σ): 1.4
- Significance level (alpha): 0.05

### Step 3: Standard Error and Z-Score Calculation
Since the population standard deviation is known, we use the Z-test formula for the z-score:

The standard error (SE) is calculated as follows:
[tex]\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{15}} \][/tex]

Using the z-score formula:
[tex]\[ z = \frac{\bar{x} - \mu}{SE} = \frac{4.75 - 4.5}{\frac{1.4}{\sqrt{15}}} \][/tex]

### Step 4: Finding the p-value
The p-value is the probability that the observed data would occur if the null hypothesis were true. It is computed based on the z-score obtained in the previous step.

### Step 5: Decision Making
We compare the p-value with the significance level (α = 0.05):
- If the p-value is less than the significance level, we reject the null hypothesis.
- If the p-value is greater than or equal to the significance level, we fail to reject the null hypothesis.

Here are the actual results from these computations:
- Z-score: 0.6916041689656102
- P-value: 0.4891859428731906
- Decision: The p-value (0.4892) is greater than the significance level (0.05).

### Final Decision:
Because the p-value is greater than the significance level, we do not have sufficient evidence to reject the null hypothesis. Therefore, we fail to reject the null hypothesis [tex]\( H₀: \mu = 4.5 \)[/tex]. Based on the data, it is reasonable to conclude that the average time teenagers spend on the phone per week is not significantly different from 4.5 hours.
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