for randomly selected adults, IQ scores are normally distributed with a standard deviation of 15. the scores of 14 randomly selected college students are listed below. use a 0.10 significance level to test the claim that the standard deviation of IQ scores of college students is less than 15. round the smaple standard deviation to three decimal places.
115, 128, 107, 109, 116, 124, 135
127, 115, 104, 118, 126, 129, 133
Test statistic: chi 2 = 5.571. Critical value: chi 2 = 7.042.
Null and alternative hypotheses:
Ho: The population standard deviation is = 15
Ha: The population standard deviation is < 15
Data given:
sample standard deviation = ? (Note: standard deviation is the square root of the variance; therefore, variance is standard deviation squared).
level of significance = 0.10
n = 14
Calculate the standard deviation from the data given.
The test statistic used in making a decision is the chi-square.
Here's the formula:
chi-square = (sample size - 1)(sample variance)/(value specified in the null hypothesis)
Note: Convert both standard deviations to variances when working with the formula.
Degrees of freedom is equal to n - 1, which is 13.
Checking a chi-square table using alpha = 0.10 with 13 degrees of freedom, find the critical value.
If the test statistic exceeds the critical value from the chi-square table, then the null is rejected and you can accept the alternative hypothesis. If the test statistic does not exceed the critical value from the table, then the null cannot be rejected and you cannot conclude a difference.
I hope this will help get you started.
To test the claim that the standard deviation of IQ scores of college students is less than 15, we can use the chi-square test with the following steps:
Step 1: Set up the null and alternative hypotheses:
Null Hypothesis (H0): The standard deviation of IQ scores of college students is equal to or greater than 15.
Alternative Hypothesis (Ha): The standard deviation of IQ scores of college students is less than 15.
Step 2: Calculate the sample standard deviation:
To find the sample standard deviation, use the formula:
Sample Standard Deviation = √[Σ(x - mean)^2 / (n - 1)]
where Σ represents the summation, x represents each individual score, mean represents the mean of the scores, and n represents the sample size.
Given the scores:
115, 128, 107, 109, 116, 124, 135, 127, 115, 104, 118, 126, 129, 133
First, calculate the mean (x̄):
x̄ = (115 + 128 + 107 + 109 + 116 + 124 + 135 + 127 + 115 + 104 + 118 + 126 + 129 + 133) / 14
x̄ = 1561 / 14
x̄ ≈ 111.5
Next, calculate the sum of squared differences:
Σ(x - mean)^2 = (115 - 111.5)^2 + (128 - 111.5)^2 + ... + (133 - 111.5)^2
Calculating the sum gives:
Σ(x - mean)^2 = 984.25
Finally, calculate the sample standard deviation:
Sample Standard Deviation = √[Σ(x - mean)^2 / (n - 1)] = √[984.25 / (14 - 1)] = √[984.25 / 13] ≈ √75.731 ≈ 8.711 (rounded to three decimal places)
Step 3: Calculate the test statistic:
The test statistic for this problem is the chi-square statistic, which can be calculated using the formula:
Chi-Square = [(n - 1) * sample variance] / population variance
where n is the sample size, the sample variance is equal to the square of the sample standard deviation, and the population variance is equal to the square of the hypothesized standard deviation (15 in this case).
Plugging in the values:
Chi-Square = [(14 - 1) * (8.711)^2] / (15^2) ≈ [13 * 75.731] / 225 ≈ 4.369
Step 4: Determine the critical value:
Since the significance level is 0.10 and the alternative hypothesis is that the standard deviation is less than 15, we need to find the critical value from the chi-square distribution table with degrees of freedom equal to (n - 1) = (14 - 1) = 13 and a significance level of 0.10.
The critical value is 19.81.
Step 5: Compare the test statistic with the critical value:
Since the test statistic value (4.369) is less than the critical value (19.81), we fail to reject the null hypothesis.
Step 6: Interpret the result:
Based on the obtained result, there is not enough evidence to support the claim that the standard deviation of IQ scores of college students is less than 15.
To test the claim that the standard deviation of IQ scores of college students is less than 15, we can conduct a hypothesis test.
The given sample only provides us with the IQ scores of 14 college students. To perform the hypothesis test, we need to calculate the sample standard deviation.
1. Calculate the sample standard deviation:
- Add up all the IQ scores: 115 + 128 + 107 + 109 + 116 + 124 + 135 + 127 + 115 + 104 + 118 + 126 + 129 + 133 = 1,688
- Calculate the sample mean: 1688 / 14 = 120.571
- Calculate the squared differences of each IQ score from the mean:
(115 - 120.571)^2 + (128 - 120.571)^2 + ... + (133 - 120.571)^2 = 14,919.571
- Divide the sum by (n-1) (where n is the number of data points) to get the sample variance: 14919.571 / (14-1) = 1114.226
- Finally, take the square root of the sample variance to get the sample standard deviation: √1114.226 ≈ 33.380 (rounded to three decimal places)
2. State the null and alternative hypotheses:
- Null hypothesis (H0): The standard deviation of IQ scores of college students is equal to or greater than 15 (σ >= 15)
- Alternative hypothesis (Ha): The standard deviation of IQ scores of college students is less than 15 (σ < 15)
3. Determine the test statistic:
- We can use the chi-square distribution to test the hypotheses.
- Calculate the test statistic using the formula: χ² = (n - 1) * (s² / σ²), where n is the sample size, s is the sample standard deviation, and σ is the hypothesized standard deviation.
- In this case, n = 14, s = 33.380, and σ = 15.
- Calculate χ²: χ² = (14 - 1) * (33.380² / 15²) ≈ 38.326
4. Determine the critical value:
- Since the alternative hypothesis states that the standard deviation is less than 15, it is a left-tailed test.
- Look for the critical value in the chi-square distribution table with (n - 1) degrees of freedom and a significance level of 0.10.
- For n - 1 = 13 degrees of freedom and a significance level of 0.10, the critical value is approximately 20.483.
5. Make a decision:
- Compare the test statistic (38.326) with the critical value (20.483).
- If the test statistic is greater than the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
- In this case, 38.326 > 20.483, so we reject the null hypothesis.
6. State the conclusion:
- Based on the sample data, there is sufficient evidence to support the claim that the standard deviation of IQ scores of college students is less than 15 at a significance level of 0.10.