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JulijaS [17]
2 years ago
9

An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

13 pounds. A consumer activist decides to test the authenticity of the claim. She follows the progress of 30 women who recently joined the weight-reduction program. She calculates the mean weight loss of these participants as 13.8 pounds with a standard deviation of 3.1 pounds. The test statistic for this hypothesis would be __________.
Mathematics
1 answer:
Sauron [17]2 years ago
6 0

The test statistic for the hypothesis would be 1.413.

Given that the participants in its new diet program lose, on average, more than 13 pounds and the mean weight loss of these participants as 13.8 pounds with a standard deviation of 3.1 pounds.

The objective is to text the advertisement's claim that participants in new diet program lose weight, on average, more than 13 pounds.

Hypothesis:

Null hypothesis:H₀:μ=13

Alternative hypothesis:Hₐ:μ>13

Here, μ be the mean weight loss of all participants.

n=30,

Degree of freedom n-1=30-1=29

\bar{x}=13.8 and s=3.1

To test the null hypothesis H₀, the value of test static would be calculated as follows:

\begin{aligned}t&=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\\ t_{29}&=\frac{13.8- 13}{\frac{3.1}{\sqrt{30}}}\\ &=\frac{0.8}{0.566}\\ &=1.413\end

Hence, the value of the test static for the hypothesis with the mean weight loss of these participants as 13.8 pounds with a standard deviation of 3.1 pounds is 1.413.

Learn more about hypothesis from here brainly.com/question/14783359

#SPJ4

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Answer:

a) P(X =16 ) = 0.1853

b) P(X \leq 12) = 0.0684

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a)

P(X =16 ) = ^{16}C_{16} * 0.90^x * ( 1 - 0.90)^{16-16}

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= 1 - [ P(X = 13) +P(X = 14) +P(X = 15) +P(X = 16) ]

= 1 - [ ^{16}C_{13} * 0.90^{13} * (1 - 0.90)^3 +^{16}C_{14} * 0.90^{14} * (1 - 0.90)^2 +^{16}C_{15} * 0.90^{15} * (1 - 0.90)^1 +^{16}C_{16} * 0.90^{16} * (1 - 0.90)^0 ]

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Second, solve for the y - intercept. Equate x to 0 which will give us,

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