# 4The null and altemate hypothe oe.The following cample Information chows the number of defectlve unta produced on the day shtft and the afternoon shtft for a sampleof four days lat month.Day2 3i0 11 14 18Day shi ftAf tornoonshift91318At the 0.010 cgnticance leve, can we conclude there are more defects produced on the day shift? HintFor the calculations, acoumethe day chtft as the first somple.State the declclon rule. (Round your answe to 2 decimal places.)b. €oneutc the voluc o oc teee 5చడంగం (Round your ancwer to a డేజియme pleccc.wV Bctaeen 0025 cnd 0.03soo o 9v2200042*4*9.0001d Whet o your dec2o1 ego/crg 3rReest

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I wwould like the solutions for parts a-d

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Step 1

Decision rule:

Reject H0 if test statistic > Critical value.

In this case, the appropriate test is paired t test,

The test statistic is t which is calculated as,

Step 2

where d-bar is the mean of the differences, µd0 is the hypothesized mean difference, sd-bar  is the standard deviation of the differences.

Step-by-step procedure to obtain the test statistic and critical value using Excel software:

• In Excel sheet, enter Day shift and Afternoon shift in different columns.
• In Data, select Data Analysis and Choose t-Test: Paired Two Sample for Means.
• In Variable 1 Range, select Day shift and in Variable 2 Range se...

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