Question

A consumer reports that the mean price of refrigerators is greater than 320. US a= 0.05.
The numbers are
333
394
344
298
336
295
352
367
313
320

Answer 1

What is the test statistic

Answer 2

What is the pvalue for 3.345

Answer 3

To determine if the mean price of refrigerators is greater than $320, we can perform a one-sample t-test. Here are the steps involved:

1. **State the Hypotheses**
- Null Hypothesis (\(H_0\)): \(\mu \leq 320\) (the mean price is less than or equal to $320)
- Alternative Hypothesis (\(H_1\)): \(\mu > 320\) (the mean price is greater than $320)

2. **Choose the Significance Level**
- \( \alpha = 0.05 \)

3. **Calculate the Sample Mean and Sample Standard Deviation**
- Let's calculate the mean and standard deviation of the given prices.

Given data:
- 333, 394, 344, 298, 336, 295, 352, 367, 313, 320

**Calculating the Mean (\(\bar{x}\)):**
\[
\bar{x} = \frac{\sum x}{n} = \frac{333 + 394 + 344 + 298 + 336 + 295 + 352 + 367 + 313 + 320}{10}
\]

**Calculating the Sum:**
\[
333 + 394 + 344 + 298 + 336 + 295 + 352 + 367 + 313 + 320 = 3850
\]
\[
\bar{x} = \frac{3850}{10} = 385
\]

**Calculating the Standard Deviation (s):**
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
\]

First, calculate \((x_i - \bar{x})^2\) for each price:

\[
\text{Data: } [333, 394, 344, 298, 336, 295, 352, 367, 313, 320]
\]
\[
(333 - 385)^2 = 2,716
\]
\[
(394 - 385)^2 = 81
\]
\[
(344 - 385)^2 = 1,681
\]
\[
(298 - 385)^2 = 7,569
\]
\[
(336 - 385)^2 = 2,401
\]
\[
(295 - 385)^2 = 8,100
\]
\[
(352 - 385)^2 = 1,089
\]
\[
(367 - 385)^2 = 324
\]
\[
(313 - 385)^2 = 5,121
\]
\[
(320 - 385)^2 = 4,225
\]


Now sum these squared differences:
\[
2,716 + 81 + 1,681 + 7,569 + 2,401 + 8,100 + 1,089 + 324 + 5,121 + 4,225 = 33,928
\]

Now plug this value into the standard deviation formula:
\[
s = \sqrt{\frac{33,928}{10-1}} = \sqrt{3,770.89} \approx 61.4
\]

4. **Calculate the t-statistic**
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{385 - 320}{61.4 / \sqrt{10}}
\]
\[
t = \frac{65}{61.4 / 3.1623} \approx \frac{65}{19.43} \approx 3.34
\]

5. **Determine the Degrees of Freedom**
\[
df = n - 1 = 10 - 1 = 9
\]

6. **Find the Critical t-value for \( \alpha = 0.05\) (one-tailed)**
Using a t-table, find \( t_{crit} \) for \( df = 9\) at \( \alpha = 0.05 \):
- \( t_{crit} \approx 1.833 \)

7. **Make a Decision**
- If \( t > t_{crit} \), we reject the null hypothesis.
- Here, \( 3.34 > 1.833\), so we reject the null hypothesis.

8. **Conclusion**
There is sufficient evidence at the 0.05 significance level to conclude that the mean price of refrigerators is greater than $320.

Answer 4

To find the test statistic, we already performed the calculations previously. However, let's present the derivation again clearly.

### Step-by-Step Calculation of the Test Statistic:

1. **Given Data**:
- Prices: \( 333, 394, 344, 298, 336, 295, 352, 367, 313, 320 \)

2. **Calculate the Sample Mean (\(\bar{x}\))**:
- \[
\bar{x} = \frac{333 + 394 + 344 + 298 + 336 + 295 + 352 + 367 + 313 + 320}{10} = \frac{3850}{10} = 385
\]

3. **Calculate the Sample Standard Deviation (s)**:
- First, we compute the sum of squared differences from the mean.
- Deviations Squared:
- \( (333 - 385)^2 = 2,716 \)
- \( (394 - 385)^2 = 81 \)
- \( (344 - 385)^2 = 1,681 \)
- \( (298 - 385)^2 = 7,569 \)
- \( (336 - 385)^2 = 2,401 \)
- \( (295 - 385)^2 = 8,100 \)
- \( (352 - 385)^2 = 1,089 \)
- \( (367 - 385)^2 = 324 \)
- \( (313 - 385)^2 = 5,121 \)
- \( (320 - 385)^2 = 4,225 \)

- Total Sum of Squared Deviations:
- \[
2,716 + 81 + 1,681 + 7,569 + 2,401 + 8,100 + 1,089 + 324 + 5,121 + 4,225 = 33,928
\]

- Sample variance:
\[
s^2 = \frac{33,928}{n - 1} = \frac{33,928}{10 - 1} = \frac{33,928}{9} \approx 3,770.89
\]

- Sample standard deviation:
\[
s = \sqrt{3,770.89} \approx 61.4
\]

4. **Calculate the Test Statistic (t)**:
- We use the formula for the t-statistic:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]
- Where:
- \(\bar{x} = 385\) (sample mean)
- \(\mu_0 = 320\) (hypothesized population mean)
- \(s \approx 61.4\) (sample standard deviation)
- \(n = 10\) (sample size)

Therefore:
\[
t = \frac{385 - 320}{61.4 / \sqrt{10}} = \frac{65}{61.4 / 3.1623} \approx \frac{65}{19.43} \approx 3.34
\]

### Conclusion:
The test statistic \( t \) is approximately **3.34**.

Answer 5

To find the p-value for a t-statistic of approximately 3.345 with 9 degrees of freedom (since the sample size \( n = 10 \), degrees of freedom \( df = n - 1 = 9 \)), we can consult a t-distribution table or use a statistical software/calculator.

For a one-tailed test with a t-statistic of 3.345 and 9 degrees of freedom:

1. **Using a t-table**:
Look up the value of 3.345 in the t-table for 9 degrees of freedom to find the corresponding tail probability.

2. **Using Statistical Software**:
Alternatively, you can calculate the p-value using statistical software or an online calculator.

Using Python, R, or any statistical software, you could use the cumulative distribution function (CDF) for the t-distribution to find the p-value. For a t-distribution:

\[
\text{p-value} = 1 - P(T \leq t) = 1 - CDF(t)
\]

In this case, if you were to use Python you might do the following:

```python
from scipy import stats

# Given values
t_statistic = 3.345
df = 9

# Calculate the p-value for the one-tailed t-test
p_value = 1 - stats.t.cdf(t_statistic, df)
print(p_value)
```

### Estimating the p-value:
From statistical tables or calculators, for \( t = 3.345 \) and \( df = 9 \):
- The p-value is approximately around **0.002** to **0.003**.

### Conclusion:
The p-value for a t-statistic of **3.345** with **9 degrees of freedom** (for a one-tailed test) is approximately **0.002** to **0.003**. This indicates a highly significant result, leading to rejection of the null hypothesis if we were using a significance level of 0.05.