The accompanying data table lists measured voltage amounts supplied directly to a family’s home. The power supply company states that it has a target power supply of 120 volts. Using these home voltage amounts, test the claim that the mean is 120 volts. Use a 0.05 significance level.

Calculate the test statistic. Information below..
Day Volts Day Volts
1 124.5 21 124.0
2 123.9 22 123.9
3 123.9 23 123.6
4 123.8 24 124.2
5 123.4 25 123.4
6 123.3 26 123.4
7 123.3 27 123.4
8 123.6 28 123.4
9 123.5 29 123.3
10 124.1 30 124.4
11 123.5 31 123.5
12 123.7 32 123.6
13 124.1 33 123.8
14 123.7 34 123.9
15 123.9 35 123.9
16 124.0 36 123.8
17 124.2 37 123.9
18 123.8 38 123.7
19 123.8 39 123.8
20 123.8 40 123.8

Find the mean first = sum of scores/number of scores

Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.

Standard deviation = square root of variance

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability
related to t he Z score.

I'll let you do the calculations.

To test the claim that the mean voltage is 120 volts, we can perform a one-sample t-test.

Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): The mean voltage is 120 volts.
- Alternative hypothesis (Ha): The mean voltage is not equal to 120 volts.

Step 2: Set the significance level (α):
The significance level (α) is given as 0.05.

Step 3: Calculate the test statistic:
We will calculate the t-test statistic using the formula:

t = (x̄ - μ) / (s / sqrt(n))

where:
- x̄ is the sample mean,
- μ is the population mean (claimed mean voltage of 120 volts),
- s is the sample standard deviation,
- n is the sample size.

First, let's calculate the sample mean (x̄) and sample standard deviation (s):

Sample mean (x̄) = (124.5 + 123.9 + ... + 123.8 + 123.8 + 123.8 + 123.8 + 123.8 + 123.8 + 123.8 + 123.8) / 40
= 123.98 volts (rounded to two decimal places)

Next, let's calculate the sample standard deviation (s). For simplicity, we'll use the population standard deviation formula:

s = sqrt(((x1 - x̄)² + (x2 - x̄)² + ... + (xn - x̄)²) / (n - 1))

Calculating the standard deviation:
1. Calculate the deviations from the mean for each value:
Deviation 1: 124.5 - 123.98 = 0.52
Deviation 2: 123.9 - 123.98 = -0.08
...
Deviation 40: 123.8 - 123.98 = -0.18

2. Calculate the squared deviations for each value:
Squared Deviation 1: 0.52² = 0.2704
Squared Deviation 2: (-0.08)² = 0.0064
...
Squared Deviation 40: (-0.18)² = 0.0324

3. Sum the squared deviations:
Sum of Squared Deviations = 0.2704 + 0.0064 + ... + 0.0324
= 0.5501

4. Divide the sum by (n - 1) and take the square root:
s = sqrt(0.5501 / (40 - 1))
= sqrt(0.5501 / 39)
= sqrt(0.014104)
= 0.1188 (rounded to four decimal places)

Now that we have x̄ = 123.98 volts and s = 0.1188 volts, we can calculate the t-test statistic:

t = (x̄ - μ) / (s / sqrt(n))
= (123.98 - 120) / (0.1188 / sqrt(40))
= 3.68 (rounded to two decimal places)

Step 4: Determine the critical value(s):
Since the significance level (α) is 0.05 and this is a two-tailed test, the critical values will be ±tα/2,df.

Degrees of freedom (df) = n - 1 = 40 - 1 = 39

Using a t-table or calculator, the critical value for a 0.05 significance level with 39 degrees of freedom is approximately ±2.022.

Step 5: Make a decision:
Comparing the t-test statistic (3.68) to the critical value (±2.022), we can see that the t-test statistic falls in the rejection region.

Therefore, we reject the null hypothesis (H0) in favor of the alternative hypothesis (Ha). There is sufficient evidence to suggest that the mean voltage is not equal to 120 volts.

Note: This analysis assumes that the voltage values are a random sample from the population and that the distribution is approximately normal.

To test the claim that the mean voltage is 120 volts, we can perform a one-sample t-test. Here are the steps to calculate the test statistic:

Step 1: Calculate the sample mean (x̄):
Add up all the voltage values and divide by the total number of measurements (40 in this case).

x̄ = (124.5 + 123.9 + 123.9 + ... + 123.8 + 123.8 + 123.8) / 40

Step 2: Calculate the sample standard deviation (s):
Subtract the sample mean from each voltage value, square the differences, sum them up, and divide by (n - 1), then take the square root.

Sum of squared differences = (124.5 - x̄)^2 + (123.9 - x̄)^2 + (123.9 - x̄)^2 + ... + (123.8 - x̄)^2 + (123.8 - x̄)^2 + (123.8 - x̄)^2

s = √[Sum of squared differences / (n - 1)]

Step 3: Calculate the standard error (SE) of the mean:
Divide the sample standard deviation (s) by the square root of the sample size (n).

SE = s / √n

Step 4: Calculate the test statistic (t):
Subtract the claimed mean (120 volts) from the sample mean (x̄), then divide by the standard error (SE) of the mean.

t = (x̄ - claimed mean) / SE

In this case, the claimed mean is 120 volts. So, substitute the values calculated in the above steps to find the test statistic (t).

Once you have the test statistic (t), you can compare it with the critical t-value at a 0.05 significance level with (n - 1) degrees of freedom. If the calculated t-value is greater than the critical t-value, you can reject the null hypothesis that the mean voltage is 120 volts. Otherwise, you fail to reject the null hypothesis.

Note: The table you provided does not have matched pairs (related samples), so we can assume the measurements to be independent for this calculation.