The process that produces Sonora Bars (a type of candy) is intended to produce bars with a mean weight of 56 gm. The process standard deviation is known to be 0.77 gm. A random sample of 49 candy bars yields a mean weight of 55.82 gm.
(a) State the hypotheses to test whether the mean is smaller than it is supposed to be.
(b) What is the test statistic?
(c) At α = .05, what is the critical value for this test?
(d) What is your conclusion?

Answer:

|z(s)| <  |z(c)|

Then  z(s) is in the acceptance region we accept H₀.

We don´t have evidence to support that candy bar mean weight is below 56 gm

Step-by-step explanation:

Goal mean     μ  =  56 gm

standard deviation   σ  = 0,77

From random sample:

mean    x  = 55,82

sample size  n  = 49

We assume a normal distribution ( is a manufacturing controlled process)

a) Hypothesis Test:

Null Hypothesis                         H₀          x   =  μ

Alternative Hypothesis             Hₐ           x  <  μ

Alternative hypothesis is telling us that the test is a one-tail test to the left

as n > 30 we use the normal distribution.

b)  z(s) =  ( x  -  μ  ) / σ / √n

z(s)  =  55.82  -  56 ) * 7 / 0,77

z(s)  = 0,18*7/0,77

z(s)  =  - 1.6363

c)  if  significance level is α  = 0.05  then  α/2 = 0,025

Then from  z table, z  score for  0,025     z(c) = -1.96

d)  Comparing  z(c)  and z(s) modules

|z(s)| <  |z(c)|

Then  z(s) is in the acceptance region we accept H₀.

We don´t have evidence to support that candy bar mean weight is below 56 gm

If no one minds I need help!

Evaluate the expression when x= -2.
х2+бх+9​

The number of homeruns hit by each player on the Bobcats baseball team is listed below.
Which measure of center best describes the approximate number of homeruns hit by a typical Bobcat player?
12, 8, 4, 3, 3, 2, 2, 2, 0

Mode

Range

Median

Mean

finding the slope to points ​

50 POINTS PLEASE HELP
A normal work day for Carl is from 8:00 a.m. to 4:30 p.m. Carl wants to figure out the degree measure that the hour hand makes while he is at work. Carl believes that if he divides 360 by 12 and multiplies the result by 8.5 he will have the correct degrees the hour hand moves. Is Carl correct? If not, choose the appropriate explanation.
A
Yes, Carl is correct. The process he takes will give him the correct degrees the hour hand moves.

B
No, Carl is incorrect. He must take the number he ends up with and subtract it from 360.

C
No, Carl is incorrect. The hour hand will go around the clock 360 degrees.

D
No, Carl is incorrect. He must first multiply 360 by 12 and then divide by 8.5.

Plsss help me, ASAP, Thank you

Can someone help me really please

2+2 brain time yessir

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A -3
B 8.4
C 0,3
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Run two separate regressions using the data in the sections of the spreadsheet titled Regression Model 1 and Regression Model 2. Use a 95% confidence level.
Fill in all the blanks below. Do not round any numbers except for when entering them in the blanks for this quiz, round these numbers to exactly 2 decimal places even if the number has only one or no decimal places. Use commas to separate thousands (e.g., 1225 should be entered as 1,225.00).
Regression Model 1
R Square expressed as a percentage rounded and carried out to two decimal places is ________% and Adjusted R Square expressed as a percentage rounded and carried out to two decimal places is _______%. Refer to the spreadsheet embedded in your Module 14 Regression document I provided you to see the calculation and format for answering this next question. The gap between R Square and Adjusted R Square expressed as a percentage rounded and carried out to two decimal places is 0.______%.
The overall regression model (was / was not) _______ significant.
Using square footage of 1,500 and mileage of 92, the model estimates that ______ hours of labor would be required, with a margin of error of _________ hours.
Regression Model 2
R Square expressed as a percentage rounded and carried out to two decimal places is _______% and Adjusted R Square expressed as a percentage rounded and carried out to two decimal places is _______%. Refer to the spreadsheet embedded in your Module 14 Regression document I provided you to see the calculation and format for answering this next question. The gap between R Square and Adjusted R Square expressed as a percentage rounded and carried out to two decimal places is 0.________%.
The overall regression model (was / was not) __________ significant.
Using square footage of 1,500, mileage of 92, and floors of 71, the model estimates that _________ hours of labor would be required, with a margin of error of ________ hours.
Model Selection
Because the R Square is (higher / lower) _________ in regression model 1 than in regression model 2 and the gap between R Square and Adjusted R Square is (smaller / larger) in regression model 1 than in regression model 2 _________ and the margin of error of the estimated labor hours is (smaller / larger) _________ in regression model 1 than in regression model 2 , we would characterize regression model (1 or 2) ___________ as being the better model.

PLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS HELP

Let A be the set of all lines in the plane. Define a relation R on A as follows. For every l1 and l2 in A, l1 R l2 ⇔ l1 is parallel to l2. (Assume that a line is parallel to itself). Which of the following is true for R?
A. R is reflexive.
B. R is symmetric.
C. R is transitive.
D. R is neither reflexive, symmetric, nor transitive.

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41 = 4
42 = 16
43 - 64

plzzzzzz help for a brianly

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1
A train trip took 4 hours. For the trip, the train traveled at 60 kilometers per hour. For the
2
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2.88 miles
4.5 miles
7.2 miles
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The best estimate close to this answer 3 11/40
A. 2 1/2
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Suppose that only 2 percent of all people are geniuses. If an IQ test indicates that Albert is a genius, but the test is only accurate 90 percent of the time, then the probability that Albert really is a genius is roughly:

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NO LINKS

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HELP PLEASE!!
Explain the error in the following exercise;
then correct it.

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