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jasenka [17]
3 years ago
6

Use technology and the given confidence level and sample data to find the confidence interval for the population mean mu . assum

e that the population does not exhibit a normal distribution. weight lost on a diet: 95 % confidence n equals 51 x overbar equals 3.0 kg s equals 5.8 kg what is the confidence interval for the population mean mu ​?
Mathematics
1 answer:
attashe74 [19]3 years ago
6 0

To solve for the confidence interval for the population mean mu, we can use the formula:

Confidence interval = x ± z * s / sqrt (n)

where x is the sample mean, s is the standard deviation, and n is the sample size

 

At 95% confidence level, the value of z is equivalent to:

z = 1.96

 

Therefore substituting the given values into the equation:

Confidence interval = 3 ± 1.96 * 5.8 / sqrt (51)

Confidence interval = 3 ± 1.59

Confidence interval = 1.41, 4.59

 

Therefore the population mean mu has an approximate range or confidence interval from 1.41 kg to 4.59 kg.

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Answer:

x = 1/2; y = 1/3

Step-by-step explanation:

2x + 3y = 2     Eq. 1

-6x + 12y = 1     Eq. 2

Eq. 1

2x + 3y = 2

2x = -3y + 2

x = -3/2 y + 1

Eq. 2

-6x + 12y = 1

De Eq. 1 sabemos que x = -3/2 y + 1

-6x + 12y = 1

-6(-3/2 y + 1) + 12y = 1

9y - 6 + 12y = 1

21y - 6 = 1

21y = 7

y = 7/21

y = 1/3

Eq. 1

2x + 3y = 2

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2x + 1 = 2

2x = 1

x = 1/2

Respuesta: x = 1/2; y = 1/3

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Find the measure of 2<br><br> A 38<br> B 92<br> C 112<br> D 136
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Find the critical value of x2 based on the given information H1 &gt;3.5 n=14 a=0.05
egoroff_w [7]

Answer:

The degrees of freedom are given by:

df = n-1=14-1=13

The significance level is \alpha=0.05 and then the critical value can be founded in th chi square table we need a quantile that accumulates 0.05 of the area in the right tail of the distribution and for this case is:

\chi^2_{\alpha}= 22.362

And if the chi square statistic is higher than the critical value we can reject the null hypothesis in favor of the alternative.

Step-by-step explanation:

We have the followign system of hypothesis:

Null hypothesis: \sigma^2 \leq 3.5

Alternative hypothesis: \sigma^2 >3.5

The degrees of freedom are given by:

df = n-1=14-1=13

The significance level is \alpha=0.05 and then the critical value can be founded in th chi square table we need a quantile that accumulates 0.05 of the area in the right tail of the distribution and for this case is:

\chi^2_{\alpha}= 22.362

And if the chi square statistic is higher than the critical value we can reject the null hypothesis in favor of the alternative.

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