The National Institute of Standards and Technology (NIST) supplies "standard copper samples" whose melting point is supposed to be exactly1084.80

C

1084.80C. To do so, NIST must check that samples that they intend to supply meet this condition. Is there reason to think that the true melting point of a new copper sample is not1084.80

C

1084.80C? To find out, NIST measures the melting point of this sample six times. Repeated measurements of the same thing vary, which is why NIST makes six measurements. These measurements are an SRS from the population of all possible measurements. This population has a Normal distribution with mean

equal to the true melting point and standard deviation=0.25

C

=0.25C.

(a) We seek evidenceagainstthe claim that=1084.80

=1084.80. What is the sampling distribution of the mean

xin many samples of six measurements of one sample if the claim is true?

The distribution is approximately Normal with the mean=1084.80

=1084.80and the standard deviation is0.102

0.102.

The distribution is approximately Normal with the mean=1084.80

=1084.80and the standard deviation is3.95

3.95.

The distribution is approximately Normal with the mean=542.40

=542.40and the standard deviation is2.204

2.204.

The distribution is approximately Normal with the mean=184.80

=184.80and the standard deviation is8

8.

Using the software of your choice, make a plot of the Normal curve for this distribution, then use that plot to select the correct plot from the options.

(b) Suppose that one of the sample means is

=1084.90

x=1084.90. Mark this value on the axis of your sketch. Another copper sample had

=1084.50

x=1084.50for six measurements. Mark this value on the axis as well, then select the plot with the correct markings.

Explain in simple language why one result is good evidence that the true melting point is1084.80

C

1084.80Cwhereas another result is evidence to doubt that1084.80

C

1084.80Cis correct; select the correct option.

Observing a value of1084.90

1084.90is not too surprising, but observing a value of1084.50

1084.50is much less likely, which provides some evidence that1084.80

C

1084.80C.

Observing a value of1084.90

1084.90is very surprising, and so is a value of1084.50

1084.50; however this provides no evidence that1084.80

C

1084.80C.

None of the options are correct.

Observing a value of1084.90

1084.90is not too surprising, but observing a value of1084.50

1084.50is not surprising either; both values provide some evidence that=1084.80

C

=1084.80C.