It is hypothesized that when homing pigeons are disoriented in a certain manner, they vrill exhibit no preference for any direction of flight after takeoff (so that the direction X should be uniformly distributed on the interval from 0° to 360°). To test this,120 pigeons are disoriented, let loose, and the direction of flight of each is recorded; the resulting data follows. Use the chi-squared test at level 0.10 to see whether the data supports the hypothesis.Direction0-<45°45-<90°90-<135° 135-<180°Frequency12171715Direction180-<225225-<270° 270-<315° 315-<360°Frequency132017A USE SALTState the appropriate hypotheses.O Ho: P1 * P2 + ... + Pg + 0.125Hại all p,'s = 0.125O Ho: P1 = P2 = ... = Pa = 0.125Hạ: no p,= 0.125O Ho: P1 * P2 + .. + Pg + 0.125Ha: at least one p, = 0.125O Ho: P1 = P2 = ... = Pa = 0.125Hạ: at least one p, + 0.125Calculate the test statistic. (Round your answer to two decimal places.)x2 = [What can be said about the P-value for the test?O P-value < 0.005O 0.005 < P-value < 0.01O 0.01 < P-value < 0.025O 0.025 < P-value < 0.05O 0.05 < P-value < 0.10O P-value > 0.10

The data shows the direction of flight after takeoff for 120 pigeons are disoriented. There are 8 groups of direction. The probability of each direction is 0.125=18.The hypothesis is, Null hypothesis: H0:p1=p2=⋯=p8=0.125 Alternative hypothesis: Ha:at least one pi≠0.125 Correct Answer: Option 4.The expected frequencies are, Direction Frequency Expected frequency 0-<45° 12 120×0.125=15 45-<90° 17 120×0.125=15 90-<135° 17 120×0.125=15 135-<180° 15 120×0.125=15 180-<225° 13 120×0.125=15 225-<270° 20 120×0.125=15 270-<315° 17 120×0.125=15 315-<360° 9 120×0.125=15 Total 120 120 The test statistic is, χ2=∑Observed-Expected2Expected=12-15215+17-15215+17-15215+15-15215+13-15215+20-15215+17-15215+9-15215=0.6000+0.2667+0.2667+0+0.2667+1.6667+0.2667+2.4000=5.7333≈5.73 Therefore, the test statistic is 5.73.The degrees of freedom is, df=k-1=8-1=7 Use the A CHI-SQUARE DISTRIBUTION TABLE: Locate the value 7 in column Degrees of freedom. Locate the value 5.73 in corresponding row for the value 7. The value 5.73 is in between 2.833 and 12.017 in the table. The area corresponding to 2.833 and 12.017 is 0.90 and 0.10. The P-value is 0.10<P-value<0.90. Therefore, the P-value > 0.10.