A researcher hypothesizes that individuals who listen to classical music will score differently from the general population on a test of spatial ability. On a standardized test of spatial ability, (u = 58). A random sample of 14 individuals who listen to classical music is given the same test. Their scores on the test are 52, 59, 63, 65, 58, 55, 62, 63, 53, 59, 57, 61, 60, 59.
a. Is this a one- or two- tailed test?
b. What are H0 and Ha for this study?
c. Compute tobt.
d. What is tcv?
e. Should H0 be rejected? What should the researcher conclude?
f. Determine the 95% confidence interval for the population mean, based on the sample mean.
a. This is a two-tailed test.
b.
- H0 (null hypothesis): Individuals who listen to classical music will score the same as the general population on the test of spatial ability.
- Ha (alternative hypothesis): Individuals who listen to classical music will score differently from the general population on the test of spatial ability.
c. To calculate tobt (t statistic), we need to calculate the sample mean, standard deviation, and perform the appropriate calculations.
Sample mean (x̄): Sum of all scores / Total number of scores
x̄ = (52 + 59 + 63 + 65 + 58 + 55 + 62 + 63 + 53 + 59 + 57 + 61 + 60 + 59) / 14 = 59.6429
Sample standard deviation (s): Square root of [(Sum of all (xi - x̄)^2) / (n - 1)]
s = sqrt[((52-59.6429)^2 + (59-59.6429)^2 + (63-59.6429)^2 + (65-59.6429)^2 + (58-59.6429)^2 + (55-59.6429)^2 + (62-59.6429)^2 + (63-59.6429)^2 + (53-59.6429)^2 + (59-59.6429)^2 + (57-59.6429)^2 + (61-59.6429)^2 + (60-59.6429)^2 + (59-59.6429)^2) / (14-1)]
Performing the calculations, s ≈ 3.0699
tobt = (x̄ - u) / (s / sqrt(n))
tobt = (59.6429 - 58) / (3.0699 / sqrt(14))
d. tcv (critical t-value) can be determined from the t-distribution table based on the significance level (α) and degrees of freedom (n-1). Since the degrees of freedom are 14-1 = 13, we need to find the t-value for α/2 with 13 degrees of freedom.
e. To determine if we should reject the null hypothesis (H0), we compare tobt and tcv. If tobt > tcv or tobt < -tcv, we reject H0.
If tobt lies within the range -tcv ≤ tobt ≤ tcv, then we fail to reject H0.
f. To calculate the 95% confidence interval, we need to find the margin of error and add/subtract it from the sample mean.
Margin of error = tcv * (s / sqrt(n))
95% confidence interval = (x̄ - margin of error, x̄ + margin of error)
a. This is a one-tailed test because the researcher has a specific hypothesis that individuals who listen to classical music will score differently from the general population on the test of spatial ability.
b. H0 (null hypothesis): There is no difference in the mean score of individuals who listen to classical music compared to the general population on the test of spatial ability.
Ha (alternative hypothesis): Individuals who listen to classical music will score differently from the general population on the test of spatial ability.
c. We need to compute tobt using the given information. To calculate tobt, we will use the formula:
tobt = (x̄ - μ) / (s / sqrt(n))
Where:
x̄ is the sample mean (average score of individuals who listen to classical music) = (52 + 59 + 63 + 65 + 58 + 55 + 62 + 63 + 53 + 59 + 57 + 61 + 60 + 59) / 14 = 59.1429
μ is the population mean (given in the question) = 58
s is the sample standard deviation, which can be calculated using the formula = sqrt(sum((xi - x̄)^2) / (n-1)) = 3.3889
n is the sample size = 14
Now we can substitute the values into the formula:
tobt = (59.1429 - 58) / (3.3889 / sqrt(14))
tobt ≈ 0.2503
d. To determine tcv (critical value), we need to look up the value from the t-distribution table. The degrees of freedom for this test are (n-1) = (14-1) = 13. Assuming a significance level of 0.05 (α = 0.05), for a one-tailed test, the critical value is approximately 1.771.
tcv ≈ 1.771
e. To determine if H0 should be rejected, we compare tobt with tcv. If tobt is greater than tcv, we reject H0; otherwise, we fail to reject H0.
In this case, tobt (0.2503) is smaller than tcv (1.771), so we fail to reject H0. Therefore, there is no sufficient evidence to conclude that individuals who listen to classical music score differently from the general population on the test of spatial ability.
f. To determine the 95% confidence interval for the population mean based on the sample mean, we can use the formula:
CI = x̄ ± (tcv * (s / sqrt(n)))
Substituting the values:
CI = 59.1429 ± (1.771 * (3.3889 / sqrt(14)))
CI ≈ 59.1429 ± 3.4647
CI ≈ (55.6782, 62.6076)
Therefore, the 95% confidence interval for the population mean, based on the sample mean, is approximately 55.6782 to 62.6076. This means that we can be 95% confident that the true population mean lies within this interval.
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though. However, I will give you a start.
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
two-tailed