Respuesta :
To determine if the t-value for the original sample falls between -t0.99 and t0.99, we need to calculate the t-value for the original sample and compare it with the critical values provided by the t-distribution table.
Given:
- Sample size (\( n \)) = 10
- Sample mean (\( \bar{x} \)) = $528.40
- Population mean (\( \mu \)) = $428.43
- Standard deviation (\( \sigma \)) = $192.00
First, we need to calculate the standard error of the mean (\( SEM \)):
\[ SEM = \frac{\sigma}{\sqrt{n}} \]
\[ SEM = \frac{192.00}{\sqrt{10}} \]
\[ SEM \approx 60.733 \]
Next, we calculate the t-value using the formula:
\[ t = \frac{\bar{x} - \mu}{SEM} \]
\[ t = \frac{528.40 - 428.43}{60.733} \]
\[ t \approx 1.647 \]
Now, we compare the calculated t-value (1.647) with the critical values -t0.99 and t0.99. We'll consult the t-distribution table or use statistical software to find these critical values. For a two-tailed test with 9 degrees of freedom (since \( n - 1 = 10 - 1 = 9 \)), the critical t-values are approximately -2.262 and 2.262 for a 0.01 significance level (corresponding to the 99% confidence interval).
Since the calculated t-value (1.647) falls between -2.262 and 2.262, we can conclude that the t-value for the original sample does fall between -t0.99 and t0.99. Therefore, at the 99% confidence level, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the population mean differs significantly from the sample mean.
Brainlist me please
Given:
- Sample size (\( n \)) = 10
- Sample mean (\( \bar{x} \)) = $528.40
- Population mean (\( \mu \)) = $428.43
- Standard deviation (\( \sigma \)) = $192.00
First, we need to calculate the standard error of the mean (\( SEM \)):
\[ SEM = \frac{\sigma}{\sqrt{n}} \]
\[ SEM = \frac{192.00}{\sqrt{10}} \]
\[ SEM \approx 60.733 \]
Next, we calculate the t-value using the formula:
\[ t = \frac{\bar{x} - \mu}{SEM} \]
\[ t = \frac{528.40 - 428.43}{60.733} \]
\[ t \approx 1.647 \]
Now, we compare the calculated t-value (1.647) with the critical values -t0.99 and t0.99. We'll consult the t-distribution table or use statistical software to find these critical values. For a two-tailed test with 9 degrees of freedom (since \( n - 1 = 10 - 1 = 9 \)), the critical t-values are approximately -2.262 and 2.262 for a 0.01 significance level (corresponding to the 99% confidence interval).
Since the calculated t-value (1.647) falls between -2.262 and 2.262, we can conclude that the t-value for the original sample does fall between -t0.99 and t0.99. Therefore, at the 99% confidence level, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the population mean differs significantly from the sample mean.
Brainlist me please
