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2. Hypothesis tests about a population mean, population standard deviation known Aa Aa Lenders tighten or loosen their standaUse the Distributions tool to help you answer the questions that follow. Normal Distribution Mean = 730 Standard Deviation =

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A hypothesis formulated helps to test whether banks have loosened their standards for issuing credit cards.If the credit card score in the last 6 months is less than avg, credit score initially,the standards have loosened as the banks have issued credit cards to low scorers.

1.Our hypothesis is

Ho u<731

Ha =u>731

Ho tests if the the avg. credit score has fallen below 731.If Ho is rejected,we can conclude that the avg credit score for issuing credit cards have risen. i.e. standards have tightened.. If Ho is not rejected, standards have loosened.(Avg. credit card scores have fallen.)

We have used \mu and not \overline{x} because our aim is to test for the avg. credit score of the population and not of a sample of the population.In conclusion, option b. is correct.

2.If the null hypothesis is true as an inequality, the sampling distribution of \overline{x} is formulated as a Normal distribution because population s.d.(g=0.76) is known. When population s.d. is not known,\overline{x} follows t-dist.

If the null hypothesis is true as a an equality, the sampling distribution of \overline{x} is approximated by a standard normal distribution with mean 731 and standard deviation=76 V100 /n.

[Since E= ].i.e. expectation of sample mean equals population mean.

3.The value of standardized test statistic is T E(T) s.e. ( Vn 717 731 -1.842 76 100

4.For a left,one- tailed hypothesis ,a standard normal distribution at 10 % level of significance,the z-value(critical value) from the table is -1.282. Therefore,our null hypothesis is rejected,if computed z-statistic is less than the tabulated z-statistic(critical value). option a. is correct. Reject H0 if  1.282.

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5.p-value corresponding to 1.842 is .0327. (look up the probability of 1.842 from std. normal distribution tables and subtract 1 from it)

6.Using the critical value approach,the Null hypothesis is rejected because critical value(-1.282) higher than computed value(-1.842).Using the p-value approach,the null hypothesis is rejected, because p value(0.0327) is less than level of significance (0.10). Therefore ,you cannot conclude that banks have loosened their standards for issuing credit cards since 2002.

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