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3241004551 [841]
3 years ago
15

A consumer believes that a certain potato chip maker is putting fewer chips in their regular bags of chips than the advertised a

mount of 12 ounces. In order to test the null hypothesis that the average chip weight is 12 ounces per bag vs. the alternative hypothesis that the average chip weight is less than 12 ounces per bag, a random sample of 49 bags were selected. The resulting data produced a p - value of 0.136.(a) At a 5% level of significance, should the null hypothesis be rejected? (Type: Yes or No):(b) At a 10% level of significance, should the null hypothesis be rejected? (Type: Yes or No): In a statistical test of hypotheses, saying that ''the evidence is statistically significant at the .05 level'' meansA. the p - value is at least .05.B. the p - value is less than .05.C. ? is more than .25.D. ?=.10.
Mathematics
1 answer:
dybincka [34]3 years ago
3 0

Answer:

a) No

b) No

c) P value is more than 0.05.  

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 12 ounces

Sample size, n = 49

P-value = 0.136

First, we design the null and the alternate hypothesis

H_{0}: \mu = 12\text{ ounces}\\H_A: \mu < 12\text{ ounces}

We use One-tailed z test to perform this hypothesis.

a) Alpha, α = 0.05

Since, p-value > α,

The null hypothesis should not be rejected. We accept the null hypothesis and reject the alternate hypothesis. We conclude that the average chip weight is 12 ounces per bag.

b) Alpha, α = 0.10

Since, p-value > α,

The null hypothesis should not be rejected. We accept the null hypothesis and reject the alternate hypothesis. We conclude that the average chip weight is 12 ounces per bag.

c) The evidence is statistically significant at the .05 level means that the p value is more than 0.05.

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Two planes start flying towards each other simultaneously from two different cities. The distance between the cities is 960 mile
ANTONII [103]

Answer:

257.5 mph

332.5 mph

Step-by-step explanation:

The initial distance between the two planes is 960 miles, while the final distance (after t=1.5 h) is 75 miles, so the total distance covered by the two planes in 1.5h is

d=960-75=885 miles

Calling v1 and v2 the velocities of the two planes, we have the following equations:

(1) v_1 = v_2 + 75 --> velocity of plane 1 exceeds velocity of plane 2 by 75 mph

(2) v_1 t+ v_2 t=885 --> the total distance covered by the two planes is 885 miles (t=1.5 h is the time, and the products v1 t and v2 t represent the distance covered by each plane)

Substituting t=1.5 h, the second equation becomes:

1.5 v_1+1.5 v_2=885\\v_1 + v_2 = 590

By substituting (1) into this last equation, we find:

(v_2+75)+v_2 = 590\\2v_2 + 75 = 590\\2v_2 = 515\\v_2=257.5

And substituting this back into eq.(1), we find

v_1 = 257.5 + 75=332.5

So, the speeds of the two planes are

257.5 mph

332.5 mph



3 0
3 years ago
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