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castortr0y [4]
3 years ago
14

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 229229 accurate orders and 7070 that were not ac

curate. a. Construct a 9090​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 9090​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: 0.2160.216less than
Mathematics
1 answer:
tiny-mole [99]3 years ago
5 0

Answer:

a) 90% Confidence interval for true proportion of orders that are not accurate

= (0.194, 0.274)

In percentage terms, (19.4%, 27.4%)

b) Since the two confidence intervals overlap, neither restaurant appears to have a significantly different percentage of orders that are not accurate.

Step-by-step explanation:

a) Confidence Interval for the population proportion of orders that are not accurate is basically an interval of range of values where the true population proportion of orders that are not accurate can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion of orders that were not accurate = 70 ÷ (70 + 229) = 0.2341

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 90% confidence interval for sample size of (70+229=299) is obtained from the z-tables because although, the population standard deviation is not known, the sample size is large enough.

Critical value = 1.645 (from the z-tables)

Standard error of the mean = σₓ = √[p(1-p)/N]

p = sample proportion of orders that are not accurate = 0.2341

n = sample size = 299

σₓ = √(0.2341×0.7659/299) = 0.02449

95% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.2341 ± (1.645 × 0.02449)

CI = 0.2341 ± 0.04028

90% CI = (0.19382, 0.27438)

90% Confidence interval = (0.194, 0.274)

b) Confidence Interval for true proportion of orders that are not accurate

For Restaurant A = (0.194, 0.274)

For Restaurant B = (0.216, 0.297)

There is overlap. For example, p = 0.235 is possible for both intervals, which means that both restaurants could have a 23.5% inaccuracy rate. If there was no overlap of the intervals, then it would be impossible for p to be the same value for both restaurants. So again, we could conclude that the restaurants could have the same inaccurate order rate due to the overlap. There just isn't enough evidence to suggest that the true percentage of orders that are not accurate for the 2 restaurants are very different.

Hope this Helps!!!

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