A study is done to find the 95% confidence interval for the difference in means between the ACT scores between two schools in a given city. One school had a random sample of 39 students had a sample mean of 25.90 with a sample standard deviation of 3.49. The second school had a random sample of 37 students had a sample mean 27.70 with a sample standard deviation of 2.47. What is the lower bound of the 95% confidence interval for the difference in means

Solution :

95% confidence interval for the difference between the means of ACT scores between two schools, is given by :

M.E. , Margin of error,

 = 9.222

s = 3.03

        = 1.4

Therefore, 95% CI = [(25.90-27.70) - 1.4 , (25.90-27.70) + 1.4]

                              = [-3.2, -0.4]

Therefore, the lower bound is -3.2

the amount you'd have after the work is $1.55

step-by-step explanation:

you subtract 95% from $31.

she will sew have sewn five dresses   and be 1/3 done with the sixth one in an hour.

step-by-step explanation:

3/4 of an hour=45 mins

lucy makes 1 dress per 11.25 minutes

45 mins+11.25 mins=56.25 mins

60 mins- 56.25 mins=3.75 mins

3.75 mins is 1/3 of of 11.25 mins

therefore she the other 1/3 would be part of the hour and she will have sewn 5 dresses and begun the sixth.

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