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ra1l [238]
3 years ago
5

Dr. Spike is a big fan of Bojangles and is particularly interested in the popularity of its celebrated wings dinner. Starting fr

om January, he would visit the local Bojangles on randomly selected days. Upon each visit, he would ask the manager how many wings dinners were sold on the previous day. By the end of March, he had visited Bojangles for a total of 31 times and thus collected 31 sale records of wings dinners. It was found that the sample mean was 26.2 sales with a sample standard deviation of 12.8. He would like to use this information to estimate the average number of wings dinner sold per day between January and March using a confidence interval. (This particular Bojangles sold 10725 wings dinners last year, but Dr. Spike does not know this.)
Which of the following facts is not relevant for evaluating whether or not a confidence interval for the mean number of wings dinners sold per day can be calculated using this data? Select one:

a. The number of sampled days is 32, which is a sufficient sample size for the confidence interval to be meaningful.
b. The days he visited were randomly selected, so the sample is representative of the population.
c. Since three sample standard deviations below the sample mean is negative (26.2-3*12.8 = -12.2), the distribution of wings dinner sales is positively skewed, and therefore not normal so the confidence interval will not be meaningful.
d. All of the above statements are relevant.
Mathematics
1 answer:
Effectus [21]3 years ago
6 0

Answer:

c

Step-by-step explanation:

If data is skewed then the upper and lower half have different amount of spread. Here there is no proof to say that data is not spread around mean or there are outliers. Therefore we can't say that the distribution of wings dinners sales in not normal.

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The probability of a success on each roll of the coin that is filipped by Dion is 0.5.

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8 0
3 years ago
In 2011, a U.S. Census report determined that 71% of college students work. A researcher thinks this percentage has changed sinc
coldgirl [10]

Answer:

Note: The full question is attached as picture below

a) Hо : p = 0.71

Ha : p ≠ 0.71

<em>p </em>= x / n

<em>p </em>= 91/110

<em>p </em>= 0.83.

1 - Pо = 1 - 0.71 = 0.29.

b) Test statistic = z

= <em>p </em>- Pо / [√Pо * (1 - Pо ) / n]

= 0.83 - 0.71 / [√(0.71 * 0.29) / 110]

= 0.12 / 0.043265

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Test statistic = 2.77

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P-value < ∝

Reject the null hypothesis.  There is sufficient evidence to support the researchers claim at the 1% significance level.

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3 years ago
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