Answer:
At a 0.01 significance level, we reject the null hypothesis and conclude that the predicted weight of 65.89 kg is significantly different from the actual weight, which could be anywhere between 53.45 kg and 78.32 kg.
Explanation:
Given the linear regression equation:
y = -107 + 1.13x
where x is the height in cm and y is the weight in kg.
To find the predicted value of y for a person with a height of 153 cm, we substitute x = 153 into the regression equation:
y = -107 + 1.13(153)
y = -107 + 172.89
y = 65.89
Therefore, the best predicted weight for an adult male who is 153 cm tall is 65.89 kg.
To check if this predicted value is statistically significant at a 0.01 significance level, we can perform a hypothesis test.
Null Hypothesis: The predicted weight for a person with a height of 153 cm is not significantly different from the actual weight.
Alternative Hypothesis: The predicted weight for a person with a height of 153 cm is significantly different from the actual weight.
We can use a t-test to test this hypothesis, with the test statistic:
t = (y_predicted - y_actual) / (s / sqrt(n))
where y_predicted is the predicted weight, y_actual is the actual weight, s is the standard error of the estimate, and n is the sample size.
The standard error of the estimate can be calculated using:
s = sqrt((1 - r^2) * Sy^2)
where Sy is the sample standard deviation of the y variable.
From the given information, we have:
Sy = 22.77 kg
r = 0.303
Therefore,
s = sqrt((1 - 0.303^2) * 22.77^2) = 20.19 kg
The sample size is n = 100.
Substituting these values into the t-test formula, we get:
t = (65.89 - y_actual) / (20.19 / sqrt(100))
t = (65.89 - y_actual) / 2.019
We want to test at a 0.01 significance level, which corresponds to a two-tailed test with a critical value of t = ±2.576 (from a t-distribution with 98 degrees of freedom, since n-2=98).
If the absolute value of t is greater than 2.576, we reject the null hypothesis and conclude that the predicted weight is significantly different from the actual weight.
Substituting t = 2.576 and t = -2.576 into the t-test formula, we get:
2.576 = (65.89 - y_actual) / 2.019
y_actual = 53.45 kg
-2.576 = (65.89 - y_actual) / 2.019
y_actual = 78.32 kg
Therefore, at a 0.01 significance level, we reject the null hypothesis and conclude that the predicted weight of 65.89 kg is significantly different from the actual weight, which could be anywhere between 53.45 kg and 78.32 kg.