Answer:
D)Yes, because the difference in the means in the actual experiment was more than two standard deviations from 0.
Step-by-step explanation:
We will test the hypothesis on the difference between means.
We have a sample 1 with mean M1=18.2 (drug group) and a sample 2 with mean M2=15.9 (no-drug group).
Then, the difference between means is:

If the standard deviation of the differences of the sample means of the two groups was 1.1 days, the t-statistic can be calculated as:

The critical value for a two tailed test with confidence of 95% (level of significance of 0.05) is t=z=1.96, assuming a large sample.
This is approximately 2 standards deviation (z=2).
The test statistict=2.09 is bigger than the critical value and lies in the rejection region, so the effect is significant. The null hypothesis would be rejected: the difference between means is significant.
I can’t solve this because there is not enough information
0.19 for one cookie and 0.59 for one doughnut
x for how much one cookie costs and y for how much one doughnut costs
6x+4y=3.5
12x+5y=5.23
solve by substitution: (elimination would work too)
4y=3.5-6x
y=0.875-1.5x
12x+5(0.875-1.5x)=5.23
12x+4.375x-7.5x=5.23
4.5x=0.855
x=0.19
now use x to solve for y:
6(0.19)+4y=3.5
1.14+4y=3.5
2.36=4y
y=0.59
Check the work: 6(0.19)+4(0.59)=3.5