on a standard test of motor coordination, a sports psychologist found that the population of a average bowlers had a mean scores of 24,with a standard deviation of 6. she tested a random sample of 30 bowling alley and found a sample mean of 26, a second random sample of 30 bowlers at Ethel's bowling alley had a mean of 18. using the criterion of p=.05 and both tails of the sampling distribution what should she conclude about each sample's representativness of the population of average bowlers?
To determine whether each sample is representative of the population of average bowlers, we can conduct hypothesis testing using the criterion of p = 0.05 (significance level) and both tails of the sampling distribution.
Let's break down the steps to perform the hypothesis testing for each sample:
Sample 1 (30 bowlers at the first bowling alley):
- Null hypothesis (H₀): The mean score of the first sample is equal to the mean score of the population (μ = 24).
- Alternative hypothesis (H₁): The mean score of the first sample is not equal to the mean score of the population (μ ≠ 24).
To test the hypothesis, we need to calculate the z-score using the formula:
z = (sample mean - population mean) / (population standard deviation / √sample size)
Substituting the values:
z = (26 - 24) / (6 / √30)
z = 2 / (6 / 5.48)
z = 2 / 0.8267
z ≈ 2.42
Now, we compare the z-score to the critical z-value corresponding to a significance level of 0.05 (two-tailed test). The critical z-value for a two-tailed test with α = 0.05 is approximately ± 1.96.
Since the calculated z-score (2.42) is greater than the critical z-value (1.96), we reject the null hypothesis. Therefore, we conclude that the first sample is not representative of the population of average bowlers.
Sample 2 (30 bowlers at Ethel's bowling alley):
- Null hypothesis (H₀): The mean score of the second sample is equal to the mean score of the population (μ = 24).
- Alternative hypothesis (H₁): The mean score of the second sample is not equal to the mean score of the population (μ ≠ 24).
Similarly, calculating the z-score:
z = (18 - 24) / (6 / √30)
z = -6 / (6 / 5.48)
z = -6 / 0.8267
z ≈ -7.26
Again, we compare the z-score (-7.26) to the critical z-value (±1.96) for a significance level of 0.05 (two-tailed test). Since the calculated z-score is well beyond the critical z-value, we reject the null hypothesis. Hence, we conclude that the second sample is also not representative of the population of average bowlers.
In summary, based on the results of the hypothesis testing, neither the first sample nor the second sample is representative of the population of average bowlers.