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Ivahew [28]
3 years ago
11

In 2011, a U.S. Census report determined that 71% of college students work. A researcher thinks this percentage has changed sinc

e then. A survey of 110 college students reported that 91 of them work. Is there evidence to support the reasearcher's claim at the 1% significance level? A normal probability plot indicates that the population is normally distributed.
Mathematics
1 answer:
coldgirl [10]3 years ago
6 0

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

= 2.77360453

Test statistic = 2.77

c) P-value

P(z > 2.77) = 2 * [1 - P(z < 2.77)] = 2 * 0.0028

P-value = 0.0056

∝ = 0.01

P-value < ∝

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

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Answer:

(A) The data set of the sample of students who has taken a calculus test is right skewed.

(B) The average of female students in the class is 66.

Step-by-step explanation:

(A)

For a left skewed distribution the Mean < Median < Mode.

For a right skewed distribution the Mean > Median > Mode.

For a symmetric distribution the Mean = Median = Mode.

Given: Mean = 78.2, Median = 67 and Mode = 67

In this case the mean of the data is more than the median and mode.

Mean = 78.2>Median = Mode = 67

Thus, the data set of the sample of students who has taken a calculus test is right skewed.

(B)

Total number of student (n) = 35

Combined average (\mu_{c}) = 70

Number of male student (n_{M}) = 20

Average of male students (\mu_{M}) = 73

Number of female student (n_{F}) = 15

Average of female students = \mu_{F}

The formula to compute the combined average is:

\mu_{c}=\frac{n_{M}\mu_{M}+n_{F}\mu_{F}}{n_{M}+n_{F}}

Compute the value of \mu_{F} as follows:

\mu_{c}=\frac{n_{M}\mu_{M}+n_{F}\mu_{F}}{n_{M}+n_{F}}\\70=\frac{(20\times73)+(15\times\mu_{F})}{20+15}\\ 70\times35=1460+(15\times\mu_{F})\\\mu_{F}=\frac{2450-1460}{15} \\=66

Thus, the average of female students in the class is 66.

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