calculating upper limit of confidence interval for normaly distributed sample with sigma unknown
$begingroup$
i have following problem,
To find the average SAT verbal score in a class, six students are sampled and their scores are 560, 610, 500, 470, 660, and 640. Assuming that students' SAT verbal scores follow normal distribution, what is the upper limit for the confidence interval of the distribution mean with confidence level 90%?
i tried calculating S^2 and then sqrt(S^2) to find sigma as it is unknown, then try to calculate upper limit of 90% confidence interval as
$$X_n + Z_p*S/sqrt(n)$$
which is equal to 573.333 + (1.645*S/2.4495)
but answe is not correct, what wrong did i do?
confidence-interval
asked Dec 2 '18 at 8:16
NourNour
234
$endgroup$
add a comment |
$begingroup$
i have following problem,
To find the average SAT verbal score in a class, six students are sampled and their scores are 560, 610, 500, 470, 660, and 640. Assuming that students' SAT verbal scores follow normal distribution, what is the upper limit for the confidence interval of the distribution mean with confidence level 90%?
i tried calculating S^2 and then sqrt(S^2) to find sigma as it is unknown, then try to calculate upper limit of 90% confidence interval as
$$X_n + Z_p*S/sqrt(n)$$
which is equal to 573.333 + (1.645*S/2.4495)
but answe is not correct, what wrong did i do?
confidence-interval
asked Dec 2 '18 at 8:16
NourNour
234
$endgroup$
add a comment |
$begingroup$
i have following problem,
To find the average SAT verbal score in a class, six students are sampled and their scores are 560, 610, 500, 470, 660, and 640. Assuming that students' SAT verbal scores follow normal distribution, what is the upper limit for the confidence interval of the distribution mean with confidence level 90%?
i tried calculating S^2 and then sqrt(S^2) to find sigma as it is unknown, then try to calculate upper limit of 90% confidence interval as
$$X_n + Z_p*S/sqrt(n)$$
which is equal to 573.333 + (1.645*S/2.4495)
but answe is not correct, what wrong did i do?
confidence-interval
asked Dec 2 '18 at 8:16
NourNour
234
$endgroup$
i have following problem,
To find the average SAT verbal score in a class, six students are sampled and their scores are 560, 610, 500, 470, 660, and 640. Assuming that students' SAT verbal scores follow normal distribution, what is the upper limit for the confidence interval of the distribution mean with confidence level 90%?
i tried calculating S^2 and then sqrt(S^2) to find sigma as it is unknown, then try to calculate upper limit of 90% confidence interval as
$$X_n + Z_p*S/sqrt(n)$$
which is equal to 573.333 + (1.645*S/2.4495)
but answe is not correct, what wrong did i do?
confidence-interval
confidence-interval
asked Dec 2 '18 at 8:16
NourNour
234
asked Dec 2 '18 at 8:16
NourNour
234
asked Dec 2 '18 at 8:16
NourNour
234
asked Dec 2 '18 at 8:16
NourNour
234
asked Dec 2 '18 at 8:16
NourNour
234
234
add a comment |
add a comment |
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