A recent study1 examines chocolate’s effects on blood vessel function in healthy people. In the randomized, double-blind, placebo-controlled study, 11 people received 46 grams (1.6 ounces) of dark chocolate (which is naturally flavonoid-rich) every day for two weeks, while a control group of 10 people received a placebo consisting of dark chocolate with low flavonoid content. Participants had their vascular health measured (by means of flow-mediated dilation) before and after the two-week study. The increase over the two-week period was measured, with larger numbers indicating greater vascular health. For the group getting the good dark chocolate, the mean increase was 1.3 with a standard deviation of 2.32, while the control group had a mean change of -0.96 with a standard deviation of 1.58. Find a 90% confidence interval for the difference in means between the two groups MC-MN , where MC represents the mean increase in flow-mediated dilation for people eating dark chocolate every day and MN represents the mean increase in flow-mediated dilation for people eating a dark chocolate substitute each day. You may assume that neither sample shows significant departures from normality.
Round your answers to two decimal places.

CI = (0.68, 3.84)

Step-by-step explanation:

We want to find the confidence interval for the difference in means between the two groups MC and MN.

where;

- MC represents the mean increase in flow-mediated dilation for people eating dark chocolate every day.

- MN represents the mean increase in flow-mediated dilation for people eating a dark chocolate substitute each day.

We are given;

Sample mean of C; x'c = 1.3

Sample mean of N; x'n = -0.96

Standard deviation of C; S_c = 2.32

Standard deviation of N; S_n = 1.58

Sample size of C; n_c = 11

Sample size of N; n_n = 10

Formula for the confidence interval for the difference in means is;

CI = (x'c - x'n) ± t√[(S_c²/n_c) + (S_n²/n_n)]

Where t is critical value at confidence level. From table attached, t at CL of 90% with DF = 10 - 1 = 9 is; t = 1.833

Thus;

CI = (1.3 - (-0.96)) ± 1.833√[(2.32²/11) + (1.58²/10)]

CI = [(1.3 + 0.96)) - 1.833√[(2.32²/11) + (1.58²/10)], [(1.3 + 0.96)) + 1.833√[(2.32²/11) + (1.58²/10)]

CI = (2.26 - 1.58), (2.26 + 1.58)

CI = (0.68, 3.84)

step-by-step explanation:

if you meant, "on the way from (3, 2) to (3, 14):

x does not change; y increases by 12.

ensure that you have copied down this problem correctly.   .

step-by-step explanation:

i found the koficent of the linear function and i foun c acordind to the strict formula y=kx+c and then i took 2 as an x and replacet it on the formula i found and then foun the y