Of 43 bank customers depositing a check 18 received some cash back Construct a 90% confidence interval for the proportion of all depositors

Question

Of 43 bank customers depositing a check 18 received some cash back Construct a 90% confidence interval for the proportion of all depositors who ask for cash back (Do not round the intermediate answers. Round your answer to 4 decimal places.) the 90% confidence interval is from to May normality of p be assumed? a. No b. Yes

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Arianna 2 weeks 2021-09-15T05:49:28+00:00 1 Answer 0

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    2021-09-15T05:51:03+00:00

    Answer:

    a) 0.419 - 1.64\sqrt{\frac{0.419(1-0.419)}{43}}=0.296

    0.419 + 1.64\sqrt{\frac{0.419(1-0.419)}{43}}=0.542

    The 90% confidence interval would be given by (0.296;0.542)

    b)  For this case we have that:

     np = 43*0.419 = 18.017 >10

     n(1-p) = 43*(1-0.419) = 24.98 >10

    And we assume independence so then we can use the normal approximation for the distribution of p.

    b. Yes.

    Step-by-step explanation:

    Notation and definitions

    X=18 number of customers with some cash back

    n=43 random sample taken

    \hat p=\frac{18}{43}=0.419 estimated proportion of customers with some cash back

    p true population proportion of customers with some cash back

    A confidence interval is “a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval”.  

    The margin of error is the range of values below and above the sample statistic in a confidence interval.  

    Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.  

    The population proportion have the following distribution

    p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

    Part a

    In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by \alpha=1-0.90=0.1 and \alpha/2 =0.05. And the critical value would be given by:

    z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64

    The confidence interval for the mean is given by the following formula:  

    \hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

    If we replace the values obtained we got:

    0.419 - 1.64\sqrt{\frac{0.419(1-0.419)}{43}}=0.296

    0.419 + 1.64\sqrt{\frac{0.419(1-0.419)}{43}}=0.542

    The 90% confidence interval would be given by (0.296;0.542)

    Part b

    For this case we have that:

     np = 43*0.419 = 18.017 >10

     n(1-p) = 43*(1-0.419) = 24.98 >10

    And we assume independence so then we can use the normal approximation for the distribution of p.

    b. Yes.

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