A family therapist states that all parents in America talk to their teenagers an average of 17 minutes per week. A psychologist decides to test this by collecting data from a sample of 9 parents on how much time they spend talking to their teenagers per week. Alpha .05, 2-tailed.
X
14
13
12
15
17
18
13
16
14
A. Give calculated score and critical score.
B. Do the psychologists data differ significantly from what the family therapist stated?
C. Is the family therapists claim an overestimate or an underestimate of actual time spent talking to children, compared to the sample size.
Calculate mean and standard deviation (SD) for your sample.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
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 related to that Z score.
A. To determine if the psychologist's data differs significantly from the family therapist's statement, we need to calculate the sample mean and the critical score.
Sample mean (X̄):
To calculate the sample mean, we sum up all the values of time spent talking to teenagers and divide it by the number of observations (n).
X̄ = (X1 + X2 + X3 + ... + X9) / n
X̄ = (14 + 13 + 12 + 15 + 17 + 18 + 13 + 16 + 14) / 9
X̄ = 132 / 9
X̄ ≈ 14.67
Calculated Score (t):
The calculated score (t) is used to determine if the psychologist's data significantly differs from the family therapist's statement. We can calculate the t-score using the formula:
t = (X̄ - μ) / (s / sqrt(n))
where μ is the population mean assumed by the family therapist (17 minutes per week), s is the sample standard deviation, and n is the sample size.
Standard deviation (s):
To calculate the sample standard deviation, we first need to calculate the variance. We sum up the squared differences between each observation and the sample mean, divide it by n-1 (to get an unbiased estimate), and then take the square root of it.
Variance (σ^2) = Σ(Xi - X̄)^2 / (n - 1)
Variance = ((14-14.67)^2 + (13-14.67)^2 + (12-14.67)^2 + (15-14.67)^2 + (17-14.67)^2 + (18-14.67)^2 + (13-14.67)^2 + (16-14.67)^2 + (14-14.67)^2) / (9-1)
Variance ≈ 4.11
s (Sample standard deviation) = sqrt(Variance)
s ≈ 2.03
t = (X̄ - μ) / (s / sqrt(n))
t = (14.67 - 17) / (2.03 / sqrt(9))
t ≈ -2.45
Critical Score:
The critical score(s) is used to compare with the calculated score to determine if the difference is statistically significant. In this case, the alpha level is 0.05, and it is a two-tailed test. Therefore, we need to find the critical value(s) from the t-distribution table with a degrees of freedom (n - 1) and an alpha level of 0.025 (0.05 / 2 for both tails).
For a sample size of 9, with a two-tailed test and an alpha of 0.025, the critical score is approximately ±2.262.
B. To determine if the psychologist's data differs significantly from what the family therapist stated, we compare the calculated score (-2.45) with the critical score (-2.262).
Since the calculated score (-2.45) falls within the critical region (-2.262), we can say that the psychologist's data does differ significantly from what the family therapist stated.
C. To determine if the family therapist's claim is an overestimate or an underestimate of the actual time spent talking to children, we compare the sample mean (14.67) with the claimed population mean (17).
Since the sample mean (14.67) is lower than the claimed population mean (17), we can say that the family therapist's claim is an overestimate of the actual time spent talking to children, compared to the given sample size.