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2 years ago

If z = -1.3, mu=80, sigma=30 What is the actual data value? Round to 2 decimal places.

1 answer:

6 0

Step-by-step explanation: This simple confidence interval calculator uses a Z statistic and sample mean (M) to generate an interval estimate of a population mean (μ).

Note: You should only use this calculator if (a) your sample size is 30 or greater; and/or (b) you know the population standard deviation (σ), and use this instead of your sample's standard deviation (an unusual situation). If your data does not meet these requirements, consider using the t statistic to generate a confidence interval.

where:

M = sample mean

Z = Z statistic determined by confidence level

sM = standard error = √(s2/n)

As you can see, to perform this calculation you need to know your sample mean, the number of items in your sample, and your sample's standard deviation (or population's standard deviation if your sample size is smaller than 30). (If you need to calculate mean and standard deviation from a set of raw scores, you can do so using our descriptive statistics tools.)

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Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, $\mu_X$ represent the population mean for Brand B and let $\mu_Y$ represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $Xbar$ = Sample mean for Brand B data = 36.9

$Ybar$ = Sample mean for Brand A data = 32.9

$n_1$ = Sample size for Brand B data = 13

$n_2$ = Sample size for Brand A data = 9

$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ , (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ]

[ $(36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ , $(36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

4 0

2 years ago

The sum of the following algebraic expression 2x + 15, 7-8x and 3x - 41 is 30 find the value of x..

Ksju [112]

$\\ \tt \longmapsto2x + 15 + 7 - 8x + 3x - 41 = 30 \\ \\ \tt \longmapsto 2x - 8x + 3x + 15 + 7 - 41 = 30 \\ \\ \tt \longmapsto - 3x - 19 = 30 \\ \\ \tt \longmapsto 3x + 19 = - 30 \\ \\ \tt \longmapsto 3x = - 30 -19 \\ \\ \tt \longmapsto 3x = - 49 \\ \\ \tt \longmapsto x = - \frac{49}{3}$

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2 years ago

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What is the answer to 7(6x-4)+2x

amm1812

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2 years ago

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-9.b = 45<br> b =<br> What is the answer I need to know the answe

BaLLatris [955]

Answer:

b=-5

Step-by-step explanation:

Divide by -9 to eliminate the -9 on -9b

(Since -9b is the same thing as -9 times b we can use the inverse of multiplication which is division to eliminate the -9 from b but we need to do it on both sides to make it balance)

-9b/-9=45/-9

b=-5

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2 years ago

Three students collected more than $93 washing cars. They used $15 to reimburse their parents for cleaning supplies. Then they d

Alenkasestr [34]

First step is find profit so lets subtract 93 and 15.
93-15=78
Finally we can divide profit by 3 students, so
78:3=26$ its the answer

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3 years ago

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