Suppose that the quality control manager for a cereal manufacturer wants to ensure that bags of cereal are being filled correctly. The equipment is calibrated to fill bags with a mean of 17 oz of cereal with a standard deviation of 0.2 oz. The quality control inspector selects a random sample of 52 boxes and finds that the mean amount of cereal for these boxes is 17.04 oz. He uses this data to conduct a one-sample z ‑test with a null hypothesis of H 0 : μ = 17 against the alternative hypothesis H 1 : μ ≠ 17 , where μ is the mean amount of cereal in each box. He calculates a z ‑score of 1.44 and a P -value of 0.1499 .
Are these results statistically significant at a significance level of 0.05?
No, these results are not statistically significant because p>0.05,
No, these results are not statistically significant because p < 0.05.
Yes, these results are statistically significant because p < 0.05.
Yes, these results are statistically significant because p > 0.05.

No, these results are not statistically significant because

p > 0.05

Step-by-step explanation:

The null hypothesis is

H0 : μ = 17

The alternative hypothesis is

H 1 : μ ≠ 17

where μ is the mean amount of cereal in each box.

The p value that he got is 0.1499. This is greater than alpha = 0.05 which is the given level of significance.

If the level of significance is lesser than the p value, we would accept the null hypothesis.

Therefore, the correct option is

No, these results are not statistically significant because p>0.05

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40

38.6°

step-by-step explanation:

resultant is 10 pound

smaller force = 8 pound

angle formed = tan⁻¹(

                        =tan⁻¹(

                          =tan⁻¹(0.8)

                        =38.6°