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Lynna [10]
4 years ago
13

Find the critical value z Subscript alpha divided by 2 that corresponds to the given confidence level. 93​%

Mathematics
1 answer:
d1i1m1o1n [39]4 years ago
4 0

Answer:

z_{{\frac{\alpha}{2}}}=z_{0.035}=1.81

Step-by-step explanation:

We are asked to compute z_{{\frac{\alpha}{2}} that corresponds to confidence level of 93%.

93\%=\frac{93}{100}=0.93

\alpha=1-0.93=0.07

\frac{\alpha}{2}=\frac{0.07}{2}

\frac{\alpha}{2}=0.035

Now, we will use normal distribution table to find the z-score that corresponds to area of 0.035.

z_{0.035}=-1.81

Therefore, the value of z_{{\frac{\alpha}{2}} is 1.81.

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Find the inverse of y=100-x^2
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Step-by-step explanation:

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he one‑sample t statistic from a sample of n = 23 observations for the two‑sided test of H 0 : μ = 15 versus H α : μ > 15 has
DedPeter [7]

Answer:

t = 2.24

The first step is calculate the degrees of freedom, on this case:  

df=n-1=23-1=22  

Since is a one side right tailed test the p value would be:  

p_v =P(t_{(22)}>2.24)=0.01776  

And for this case we can conclude that:

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And we will reject the null hypothesis at \alpha=0.025 since p_v < \alpha

Step-by-step explanation:

Data given and notation  

\bar X represent the mean height for the sample  

s represent the sample standard deviation

n=23 sample size  

\mu_o =15 represent the value that we want to test

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

For this case the statistic is given:

t = 2.24

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=23-1=22  

Since is a one side right tailed test the p value would be:  

p_v =P(t_{(22)}>2.24)=0.01776  

And for this case we can conclude that:

0.01 < p_v < 0.025

And we will reject the null hypothesis at \alpha=0.025 since p_v < \alpha

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3 years ago
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