Construct a 90% confidence interval estimate of the difference between the mean IQ score for those with a low lead level and the mean IQ score for those with a high lead level. (5 marks)
Low Lead Level: n=78, x ̅=92.88462, s=15.34451
High Lead Level: n=21, x ̅=86.90476, s=8.988352

Question

Construct a 90% confidence interval estimate of the difference between the mean IQ score for those with a low lead level and the mean IQ score for those with a high lead level. (5 marks)
Low Lead Level: 	n=78,   x ̅=92.88462,   s=15.34451
High Lead Level: 	n=21,   x ̅=86.90476,   s=8.988352

kropot72 · Answer

The sample size of n = 21 for the population with high lead level is less than 30, therefore the difference in sampling means cannot be considered to be approximately normal. However do you have the formula for calculating the required confidence interval?

kropot72 · Answer

@JacquelineC Are you there?

anonymous · Answer

I think they are considered normal distributions

anonymous · Answer

Im taking an online course and I'm not sure which formula to use from the textbook for this type of confidence interval

kropot72 · Answer

Please wait a few minutes and I will post the formula.

anonymous · Answer

I think there are two different formulas when the mean is known or the mean is unknown

kropot72 · Answer

$$(\bar{x}_{1}-\bar{x}_{2})-1.645\sqrt{\frac{s _{1}^{2}}{n _{1}}+\frac{s _{1}^{2}}{n _{2}}}$$
This formula gives the lower value of the required confidence interval. To find the upper value, just use the same formula but change the sign in front of 1.645 to plus.

anonymous · Answer

what is the 1.645 for?

kropot72 · Answer

1.645 is the value of z that is needed to find a 90% confidence interval. For example, if a 95% confidence interval was needed the value of z would be 1.96.

anonymous · Answer

Oh okay

anonymous · Answer

(x1-x2)  then subtract the rest of the equation?