In a 3M Privacy Filters poll, 806 adults were asked to identify their favorite seat when they fly, and 492 of them chose a window seat. Use a .01 significance level to test the claim that the majority of adults prefer a window seat when they fly.

Step-by-step explanation:

If majority of adults prefer a window seat when they fly, it means that more than half or more than 0.5 of the population prefer a window seat when they fly.

We would set up the hypothesis test.

For the null hypothesis,

p = 0.45

For the alternative hypothesis,

p > 0.5

Considering the population proportion, probability of success, p = 0.5

q = probability of failure = 1 - p

q = 1 - 0.5 = 0.5

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 492

n = number of samples = 806

P = 492/806 = 0.61

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.61 - 0.5)/√(0.5 × 0.5)/806 = 6.24

From the normal distribution table, the p value < 0.00001

Since alpha, 0.01 < than the p value, then we would reject the null hypothesis. Therefore, at a 1% significance level, we can conclude that the majority of adults prefer a window seat when they fly.

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.