Please see attached statistical Inference problem

Statistical Inference:

Market Research Inc. (MRI) is in the business of measuring magazine readership and product usage. EKIN Inc., a sporting goods manufacturer, wishes to use MRI to judge the effectiveness of their advertising in Sports Illustrated on the sales of their sporting equipment. To carry out this objective MRI takes an initial random sample of 1,000 households. The results from a random sample of 1,000 households in the USA are as follows:

ISSUE

CATEGORY

NUMBER OF HOUSEHOLDS

NON READER

900

READER

100

A reader household is defined as a household in which at least one of the last 4 issues of Sports Illustrated was read. A nonreader household is defined as a household in which none of the last 4 issues of Sports Illustrated was read.

The sample indicated that the median, mean and standard deviation of the EKIN sporting equipment annual expenditures for the 100 reader households are $47, $95.50 and $50 respectively.

Find a 95% confidence interval for the mean annual expenditures for all reader households in the USA.

The original data on the 100 reader households indicate that a normal distribution is not a valid assumption for the probability distribution of EKIN sporting equipment annual expenditures. Yet, the fact that the normal distribution assumption is invalid does not affect the validity of the confidence interval for the mean annual expenditures for all reader households in the USA, which is based on the normal distribution. Explain why this is true.

Suppose that there are 120 million households in the United States. Estimate the total number of reader households in the United States and obtain a 95% confidence interval for this number.

MRI wants the sampling error resulting from estimating the mean annual expenditure for all reader households in the USA to be less than $5, with 95% probability. Using the information provided in the initial sample, determine how many additional households from the US population must be sampled in order to satisfy this requirement. (NOTE: It is impossible for MRI to sample reader households. It can only take a random sample of households from the general USA population).

What famous sporting goods manufacturer does EKIN, Inc. refer to?

TABLE 1 NORMAL CURVE AREAS

AREAS UNDER THE STANDARD NORMAL CURVE FROM 0 TO Z

Z	0	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0.0	0.0000	0.0040	0.0080	0.0120	0.0160	0.0199	0.0239	0.0279	0.0319	0.0359
0.1	0.0398	0.0438	0.0478	0.0517	0.0557	0.0596	0.0636	0.0675	0.0714	0.0753
0.2	0.0793	0.0832	0.0871	0.0910	0.0948	0.0987	0.1026	0.1064	0.1103	0.1141
0.3	0.1179	0.1217	0.1255	0.1293	0.1331	0.1368	0.1406	0.1443	0.1480	0.1517
0.4	0.1554	0.1591	0.1628	0.1664	0.1700	0.1736	0.1772	0.1808	0.1844	0.1879
0.5	0.1915	0.1950	0.1985	0.2019	0.2054	0.2088	0.2123	0.2157	0.2190	0.2224
0.6	0.2257	0.2291	0.2324	0.2357	0.2389	0.2422	0.2454	0.2486	0.2517	0.2549
0.7	0.2580	0.2611	0.2642	0.2673	0.2704	0.2734	0.2764	0.2794	0.2823	0.2852
0.8	0.2881	0.2910	0.2939	0.2967	0.2995	0.3023	0.3051	0.3078	0.3106	0.3133
0.9	0.3159	0.3186	0.3212	0.3238	0.3264	0.3289	0.3315	0.3340	0.3365	0.3389
1.0	0.3413	0.3438	0.3461	0.3485	0.3508	0.3531	0.3554	0.3577	0.3599	0.3621
1.1	0.3643	0.3665	0.3686	0.3708	0.3729	0.3749	0.3770	0.3790	0.3810	0.3830
1.2	0.3849	0.3869	0.3888	0.3907	0.3925	0.3944	0.3962	0.3980	0.3997	0.4015
1.3	0.4032	0.4049	0.4066	0.4082	0.4099	0.4115	0.4131	0.4147	0.4162	0.4177
1.4	0.4192	0.4207	0.4222	0.4236	0.4251	0.4265	0.4279	0.4292	0.4306	0.4319
1.5	0.4332	0.4345	0.4357	0.4370	0.4382	0.4394	0.4406	0.4418	0.4429	0.4441
1.6	0.4452	0.4463	0.4474	0.4484	0.4495	0.4505	0.4515	0.4525	0.4535	0.4545
1.7	0.4554	0.4564	0.4573	0.4582	0.4591	0.4599	0.4608	0.4616	0.4625	0.4633
1.8	0.4641	0.4649	0.4656	0.4664	0.4671	0.4678	0.4686	0.4693	0.4699	0.4706
1.9	0.4713	0.4719	0.4726	0.4732	0.4738	0.4744	0.4750	0.4756	0.4761	0.4767
2.0	0.4772	0.4778	0.4783	0.4788	0.4793	0.4798	0.4803	0.4808	0.4812	0.4817
2.1	0.4821	0.4826	0.4830	0.4834	0.4838	0.4842	0.4846	0.4850	0.4854	0.4857
2.2	0.4861	0.4864	0.4868	0.4871	0.4875	0.4878	0.4881	0.4884	0.4887	0.4890
2.3	0.4893	0.4896	0.4898	0.4901	0.4904	0.4906	0.4909	0.4911	0.4913	0.4916
2.4	0.4918	0.4920	0.4922	0.4925	0.4927	0.4929	0.4931	0.4932	0.4934	0.4936
2.5	0.4938	0.4940	0.4941	0.4943	0.4945	0.4946	0.4948	0.4949	0.4951	0.4952
2.6	0.4953	0.4955	0.4956	0.4957	0.4959	0.4960	0.4961	0.4962	0.4963	0.4964
2.7	0.4965	0.4966	0.4967	0.4968	0.4969	0.4970	0.4971	0.4972	0.4973	0.4974
2.8	0.4974	0.4975	0.4976	0.4977	0.4977	0.4978	0.4979	0.4979	0.4980	0.4981
2.9	0.4981	0.4982	0.4982	0.4983	0.4984	0.4984	0.4985	0.4985	0.4986	0.4986
3.0	0.4987	0.4987	0.4987	0.4988	0.4988	0.4989	0.4989	0.4989	0.4990	0.4990

TABLE 2 THE STUDENT’S T DISTRIBUTION

VALUES OF THE 97.5th PERCENTILE

DEGREES OF FREEDOM.	t.025
1	12.706
2	4.303
3	3.182
4	2.776
5	2.571
6	2.447
7	2.365
8	2.306
9	2.262
10	2.228
11	2.201
12	2.179
13	2.160
14	2.145
15	2.131
16	2.120
17	2.110
18	2.101
19	2.093
20	2.086
21	2.080
22	2.074
23	2.069
24	2.064
25	2.060
26	2.056
27	2.052
28	2.048
29	2.045
infinity	1.960

Please see attached statistical Inference problem 1