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stich3
8 days ago
7

A bicycle factory installs about 400 tires per day. Tires are installed Monday through Friday for 8 hours per day. The manager o

f the factory estimates the number of properly installed tires per week using the process below.
In the first hour of work on Monday, 49 out of 50 tires were properly installed.

So, there were about 8(49) = 392 properly installed tires on Monday.

Considering 5 working days per week, 5(392) = 1,960 is the number of tires properly installed in one week.

Which best explains the validity of the results?
Mathematics
2 answers:
Inessa [9K]8 days ago
8 0
The findings are probably not valid since it is improbable that the tires fitted in just one hour on Monday accurately reflect the whole population of tires.Step-by-step explanation: Out of 8 hours, do you truly believe that one hour could stand for all? This is the reason the results are likely invalid (I also participated in the quiz).
PIT_PIT [9.1K]8 days ago
5 0
The findings are probably not valid since it is improbable that the tires fitted in just one hour on Monday accurately reflect the whole population of tires.
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A new weight-watching company, Weight Reducers International, advertises that those who join will lose an average of 10 pounds a
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Answer:

The value calculated is Z = 2.53, which exceeds 1.96 at a significance level of 0.05

This leads to the rejection of the null hypothesis H₀

Thus, we accept the alternative hypothesis

This implies that individuals participating in Weight Reducers will lose more than 10 pounds

Step-by-step explanation:

Step(i):-

The size of the random sample 'n' = 50

The new weight-loss program, Weight Reducers International, claims that members will shed an average of 10 pounds within the initial two weeks, with a standard deviation of 2.8 pounds.

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Step(ii):-

Null hypothesis: H₀: μ < 10

Alternative hypothesis: H₁: μ > 10

Test statistic calculation

z= \frac{x^{-} - mean}{\frac{S.D}{\sqrt{n} } }

z= \frac{9 - 10}{\frac{2.8}{\sqrt{50} } } = \frac{-1}{0.395} = 2.53

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In this case, the critical value Z is 1.96 at the 0.05 significance level

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Calculated Z value of 2.53 is greater than 1.96 at a significance level of 0.05

Consequently, we reject the null hypothesis H₀

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We determine that individuals who sign up for Weight Reducers will experience a weight loss exceeding 10 pounds

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Hope this helps!

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28 days ago
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