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Create a 10 pages page paper that discusses business statistics. The confidence interval for population mean could be calculated with the help of Standard Normal Variate. 95% of the area under the sta

Create a 10 pages page paper that discusses business statistics. The confidence interval for population mean could be calculated with the help of Standard Normal Variate. 95% of the area under the standard normal curve lies within the range -1.96 ≤ z ≤ 1.96, where, z = (x - μ)/ (σ/ √n).

So 90% confidence interval for the proportion of people in the population who feel insecure about their jobs could be represented as (p ± 1.64 S.E) since the area of the standard normal curve falling between ± 1.64 is 90%.

Given that, S. E. = √(pq)/ n = √(0.42 x 0.58)/ 165 = 0.038423, the 90% confidence interval about the proportion of telephone operators in UK call centres being insecure about their jobs is, [0.42 ± (1.64 x 0.038423)] = (0.356985, 0.483015).

15. The information could be represented in the form of a joint-distribution table between variables ‘Age’ and ‘Media’. However, prior to conducting joint probability distribution, the information needs to be modified as follows –

Each of the cell values needs to be divided by the total sample size to figure out the proportion of the total sample that each of them represents. The resultant table will be the joint probability distribution one.

Given that each cell represents the probability of intersection between any two values of the variables, the required calculation could be made. So, the probability of a person aged above 45 years will recall the product given that it was advertised on Television could be represented as,

19. The formula for exponential smoothing is Y (t)* = Y (t-1)* + w [Y (t) – Y (t-1)*], where w = 0.8. For simplicity, it is assumed that Y (t-1)* is the observed value for the previous year, for t = 1. Thus, the predicted value for t = 6 could be yielded as follows,

To figure out the p-value for a two-tailed test, the area above 2.66298 and below -2.66298 has to be calculated. Given that the total area under the standard normal curve is equal to 1, the percentage falling under the required area is,

Assuming the level of significance to be.

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