A random sample of 50 students has a test score average with a standard deviation of 10.9. Find
the margin of error if c = 0.90.
To find the margin of error, we first need to determine the critical value associated with a confidence level of 0.90 (c = 0.90). The critical value is obtained from the standard normal distribution or the t-distribution, depending on the sample size.
Since the sample size is 50, which is relatively large, we can use the standard normal distribution. The critical value associated with a confidence level of 0.90 is found by subtracting the significance level from 1 and dividing it by 2 (since we're looking for a two-tailed test). In this case, (1 - 0.90) / 2 = 0.05 / 2 = 0.025.
Using a standard normal distribution table or calculator, we can find that the critical value corresponding to a cumulative probability of 0.975 is approximately 1.96.
Next, we calculate the margin of error using the formula:
Margin of Error = Critical Value * Standard Error
The standard error is the standard deviation divided by the square root of the sample size. In this case, the standard error = 10.9 / √50.
Now we can substitute these values into the margin of error formula:
Margin of Error = 1.96 * (10.9 / √50)
Calculating this expression will give us the margin of error for a sample size of 50 and a confidence level of 0.90.
Margin of error = + or - 1.645 (sd/√n)
Substitute the values given and determine the margin of error.
(The value of 1.645 was found using a z-table for 90% confidence.)
I hope this helps.