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An officer states that the average fine levied b
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<p>Null Hypothesis: u<= 200.</p><p>Alternative Hypothesis: u>XXX</p><p>We assume XXX XXXX Hypothesis XX be XXXX XXX try XX either XXXXXX it or fail XX reject it.</p><p>To XXXXXXX XXXX test, XX XXXX the standard deviation XX the sample means XXXXX XX XXXXX XX the XXXXXXX:</p><p>(XXXXXX standard XXXXXXXXX)/sqrt(sample size) = 16/XXXX(10) = X.XX</p><p>XXXX we calculate XXX t-XXXXX: (XXXXXX XXXX - population XXXX)/(standard deviation XX XXXXXX XXXX) =</p><p>(XXX - XXX)/X.XX = X.XXX</p><p>XX need to XXXX the probability XX obtaining a t-score of X.XXX or higher, so XX use XXXXXX a table of p-values for a t-distribution or a XXXXXXXXXX's tcdf() function.</p><p>For either of these XXXXXXX, XX need XXX degrees of freedom which equals sample XXXX - 1 = XX - 1 = X.</p><p>XXXXX a table, XXXX XXXX the p-XXXXX for XXX t-score XXX degrees XX freedom.</p><p>XXX a calculator (XXXX XX Ti-84), XXX XXXXXXXX tcdf(1.976, X*10^99,9) = X.XXXX = p-value.</p><p>XXXXX XXX p-XXXXX, 0.XXXX, is XXXX than XXX significance level XX 0.XX, XX can XXXXXX XXX XXXX. XXXXX is significant XXXXXXXX XXXX the average XXXX is XXXXXX XXXX $XXX.</p><p>Note: we XXXXXX accept a hypothesis, only reject or XXXX XX XXXXXX. Hope this helps :)</p>">
