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Do bathroom scales tend to underestimate a person’s true weight? A 150 lb test weight was placed on each of 50 bathroom scales. The readings on 28 of the scales were too light, and the readings on the other 22 were too heavy. Can you conclude that more than half of bathroom scales underestimate weight? Find the P-value and state a conclusion. The P-value is . Round the answer to four decimal places. We cannot conclude that more than half of bathroom scales underestimate weight.

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Answer:

the P-value = 0.1977

conclusion: Since the p-value is not small;

we reject null hypothesis

hence, the data does not support claim that more than half of the bathroom scales underestimate weight.

Step-by-step explanation:

 Given the data in the question;

let X represent the number of successes ( the weight is underestimated ) in n independent Bernoulli trials, each with the success probability p

X - Bin( n,p).

so

Null hypothesis             H₀ : p ≤ 0.5

Alternative hypothesis Hₐ : p > 0.5

now, compute np₀ and n( 1 - p₀ )

np₀ = (50)(0.5)

np₀ = 25

n( 1-p₀ ) = (50)(1 - 0.5)

n( 1-p₀ ) = 25

we can see that both values are greater than 10.

∴ the sample proportion is approximately normally distributed;

P" - N ( p₀, [tex]\frac{p_{0}(1-p_0)}{n}[/tex] )

sample proportion p" = 28/50 = 0.56

standard deviation will be;

σ[tex]_{p"[/tex] = √( p₀(1-p₀) / n )

σ[tex]_{p"[/tex] = √( 0.5(1-0.5) / 50 )

σ[tex]_{p"[/tex] = √( 0.25 / 50 )

σ[tex]_{p"[/tex] = 0.07071

Next is the Z-score

z = p" - p₀ / σ[tex]_{p"[/tex]

z =  0.56-0.5 / 0.07071

z = 0.06 / 0.07071

z = 0.85

from table,

the probability that a standard normal random variable takes on a value greater than 0.85 is approximately 0.1977

Therefore, the P-value = 0.1977

conclusion: Since the p-value is not small;

we reject null hypothesis

hence, the data does not support claim that more than half of the bathroom scales underestimate weight.

p = 0.197663

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